NCERT Solutions for Class 11 Chemistry chapter 2 Structure of Atom PDF – eSaral
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### Download NCERT Solutions for Class 11 Chemistry chapter 2 Structure of Atom PDF

Question 1.Figure the quantity of electrons which will together measure one gram.

(b) Compute the mass and charge of one mole of electrons.

Solution: (a) We know : Mass of an electron $=9.10 \times 10^{-31} \mathrm{~kg}$

$\therefore$ Quantity of electrons that measure $9.10 \times 10^{-31} \mathrm{~kg}=1$

$\therefore$ Quantity of electrons that will measure $1 \mathrm{~g}\left(1 \times 10^{-3} \mathrm{~kg}\right)$

$=\frac{1}{9.1 \times 10^{-31} \mathrm{~kg}} \times\left(1 \times 10^{-3} \mathrm{~kg}\right)$

$=0.109 \times 10^{-3+31}$

$=0.1098 \times 10^{28}$

$=1.098 \times 10^{27}$

(b) (i) We know: Mass of one electron $=9.1 \times 10^{-31} \mathrm{~kg}$

Mass of one mole of electron $=9.1 \times 10^{-31} \mathrm{~kg} \times \mathrm{N}_{\mathrm{A}}=(6.023 \times$

$\left.10^{23}\right) \times\left(9.10 \times 10^{-31} \mathrm{~kg}\right)$

$=5.48 \times 10^{-7} \mathrm{~kg}$

Or

We know Mass of one electron $=9.1 \times 10^{-28} \mathrm{~g}$

The formula used: Mass of one mole of electron

$=$ Mass of one electron $\times \mathrm{N}_{\mathrm{A}}$

$=9.1 \times 10^{-28} \mathrm{~g} \times \mathrm{N}_{\mathrm{A}}$

$=\left(9.1 \times 10^{-28} \mathrm{~g}\right) \times\left(6.023 \times 10^{23}\right)$

$=5.48 \times 10^{-4} \mathrm{~g}$

We know: Charge on one electron $=1.6 \times 10^{-19}$ coulomb

Formula: Charge on one mole of electron $=$ Charge on one electron $\times \mathrm{N}_{\mathrm{A}}$

$=\left(1.6 \times 10^{-19} \mathrm{C}\right) \times \mathrm{N}_{\mathrm{A}}$

$=\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(6.023 \times 10^{23}\right)$

$=9.65 \times 10^{4} \mathrm{C}$

Question 2. (I) Calculate the total number of electrons present in one mole of methane.

(II) Find (a) the total number and (b) the total mass of neutrons in $7 \mathrm{mg}$ of $14 \mathrm{c}$.

(Assume that mass of a neutron $=1.67 \times 10^{-27} \mathrm{~kg}$ ).

(III) Find (a) the total number and (b) the total mass of protons in $34 \mathrm{mg}$ of $\mathrm{NH}_{3}$ at STP.

Will the answer change if the temperature and pressure are changed?

Solution: (I) We know : the atomic number of carbon $=6$

The atomic number of hydrogen $=1$

Number of electrons present in 1 molecule of methane $\left(\mathrm{CH}_{4}\right)$

$1(6)+4(1)=10$

Formula used: Number of electrons present in 1 mole of methane

$=$ Number of electrons present in 1 molecule of methane $\times\left(N_{A}\right)$

By putting the above values the formula

$=$ Number of electrons present in 1 molecule of methane $\times 6.023 \times$ $10^{23}$

$=10 \times 6.023 \times 10^{23}=6.023 \times 10^{24}$

(II)

(a) Number of atoms of $14_{C}$ in 1 mole $=6.023 \times 10^{23}$

Since 1 atom of $14_{c}$ contains

Neutrons $=(14-6)=8$

The number of neutrons in $14 \mathrm{~g}$ of $14_{\mathrm{C}}=N_{\mathrm{A}} \times 8=6.023 \times 10^{23} \times 8$

Or, $14 \mathrm{~g}$ of $14_{\mathrm{C}}$ contains $\left(6.023 \times 10^{23} \times 8\right)$ neutrons.

Number of neutrons in $7 \mathrm{mg}\left(7 \times 10^{-3} \mathrm{gram}\right)$

$=\frac{6.023 \times 10^{23} \times 8 \times 7 \times 10^{-3}}{14}$

$=2.40 \times 10^{21}$

(b) (x) The mass of one neutron $=1.67 \times 10^{-24} \mathrm{~g}$

The mass of total neutrons in $7 \mathrm{~g}$ of $14_{\mathrm{C}}$

$=\left(2.4092 \times 10^{21}\right) \times\left(1.67 \times 10^{-24} \mathrm{~g}\right)$

$=4.02 \times 10^{-3} \mathrm{~g}$

(y) The mass of one neutron $=1.67 \times 10^{-27} \mathrm{~kg}$

The mass of total neutrons in $7 \mathrm{~g}$ of $14 \mathrm{c}$

$=\left(2.4092 \times 10^{21}\right)\left(1.67 \times 10^{-27} \mathrm{~kg}\right)$

$=4.0352 \times 10^{-6} \mathrm{~kg}$

(III)

(a) We know: 1 mole of $\mathrm{NH}_{3}=1(14)+3(1) \mathrm{g}$ of $\mathrm{NH}_{3}$

$=17 \mathrm{~g}$ of $\mathrm{NH}_{3}$

$=6.023 \times 10^{23}$ Molecules of $\mathrm{NH}_{3}$

Total number of protons present in 1 molecule of $\mathrm{NH}_{3}$

$=1(7)+3(1)$

$=10$

Number of protons in $6.023 \times 10^{23}$ molecules of $\mathrm{NH}_{3}$

$=\left(6.023 \times 10^{23}\right)(10)$

$=6.023 \times 10^{24}$

$\Rightarrow 17 \mathrm{~g}$ of $\mathrm{NH}_{3}$

Number of protons in $34 \times 10^{-3} \mathrm{~g}$ of $\mathrm{NH}_{3}$

$=\frac{6.023 \times 10^{24} \times 34 \times 10^{-3}}{17}$

$=1.2046 \times 10^{22}$ Protons

(b)(x) We know: Mass of one proton $=1.67 \times 10^{-24} \mathrm{~g}$

The total mass of protons in $34 \mathrm{mg}\left(34 \times 10^{-3} \mathrm{~g}\right.$ ) of

$\mathrm{NH}_{3}$

$=\frac{\left(1.67 \times 10^{-24} \mathrm{~g}\right) \times 6.023 \times 10^{24} \times 34 \times 10^{-3} g}{17}$

$=\left(1.67 \times 10^{-24} \mathrm{~g}\right)\left(1.2046 \times 10^{22}\right)$

$=2.01 \times 10^{-2} \mathrm{~kg}$

(y) $\quad$ Mass of one proton $=1.67 \times 10^{-27} \mathrm{~kg}$

The total mass of protons in $34 \mathrm{mg}$ of $\mathrm{NH}_{3}$

$=\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(1.2046 \times 10^{22}\right)$

$=2.0176 \times 10^{-5} \mathrm{~kg}$

The quantity of protons, electrons, and neutrons in a particle is free of temperature what’s more, weight conditions. Subsequently, the acquired qualities will stay unaltered if the temperature and weight are changed.

Question 3. What number of neutrons and protons are there in the following nuclei?

${ }_{6}^{13} \mathrm{C},{ }_{8}^{16} \mathrm{O},{ }_{12}^{24} \mathrm{Mg},{ }_{16}^{56} \mathrm{Fe},{ }_{38}^{88} \mathrm{Sr}$

Solution: A symbolic representation of an atom ${ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{x}$ where $\mathrm{A}$ represents Atomic mass (number of proton and neutron) and $Z$ atomic number(number of proton )

${ }_{6}^{13} \mathrm{C}:$

Given: Atomic mass $(\mathrm{A})=13$

Atomic number $(\mathrm{Z})=$ Number of protons $=6$

Formula:Number of neutrons $=$ (Atomic mass)-(Atomic number)

$=13-6=7$

${ }_{8}^{16} \mathrm{O}:$

Given: Atomic mass $(\mathrm{A})=16$

Atomic number $(\mathrm{Z})=8$

Number of protons $=8$

Formula: Number of neutrons $=$ (Atomic mass)- (Atomic number)

$=16-8=8$

${ }_{12}^{24} \mathrm{Mg}:$

Given: Atomic mass $(\mathrm{A})=24$

Atomic number $(\mathrm{Z})=$ Number of protons $=12$

Formula: Number of neutrons $=$ (Atomic mass) – (Atomic number)

$=24-12=12$

${ }_{16}^{56} \mathrm{Fe}$ :

Given: Atomic mass $(\mathrm{A})=56$

Atomic number $(\mathrm{Z})=$ Number of protons $=26$

Formula: Number of neutrons $=$ (Atomic mass)-(Atomic number)

$=56-26=30$

${ }_{38}^{88} \mathrm{Sr}:$

Given: Atomic $\operatorname{mass}(\mathrm{A})=88$

Atomic number $(\mathrm{Z})=$ Number of protons $=38$

Formula: Number of neutrons $=$ (Atomic mass)-(Atomic number)

$=88-38=50$

Question 4. Compose the total image for the atom with the given nuclear number (Z) and Atomic mass (A)

(I) $\mathrm{Z}=17, \mathrm{~A}=35$

(II) $\mathrm{Z}=92, \mathrm{~A}=233$

(III) $\quad \mathrm{Z}=4, \mathrm{~A}=9$

Solution: Symbolic representation of an atom ${ }_{z}^{\mathrm{A}} \mathrm{x}$

(i) $\quad \mathrm{Z}=17, \mathrm{~A}=35:{ }_{17}^{35} \mathrm{C}$

(ii) $\mathrm{Z}=92, \mathrm{~A}=233:{ }_{92}^{233} \mathrm{U}$

(iii) $\mathrm{Z}=4, \mathrm{~A}=9:{ }_{4}^{9} \mathrm{Be}$

Question 5. Yellow light radiated from a sodium light has a wavelength $(\lambda)$ of $580 \mathrm{~nm}$. Ascertain the frequency ( $v$ ) and wave number $(\bar{v})$ of the yellow light.

Solution: given in question: $\mathrm{c}=$ velocity of light in vacuum $=3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}$

the wavelength of yellow light $(\lambda)=580 \mathrm{~nm}=580 \times 10^{-9} \mathrm{~m} / \mathrm{s}$

the formula used : wavelength $(\lambda)$

$=\frac{\text { velocity of light }(\mathrm{c})}{\text { frequency }(\mathrm{v})}$

From the expression we get,

$v=\frac{c}{\lambda} \quad \ldots \ldots \ldots$. $(1)$

Substituting the values in expression (1):

$v=\frac{3 \times 10^{8}}{580 \times 10^{-9}}=5.17 \times 10^{14} \mathrm{~s}^{-1}$

Thus, the frequency of yellow light emitted from the sodium lamp

$=5.17 \times 10^{14} \mathrm{~s}^{-1}$

The wave number of yellow light $\overline{(v)}=\frac{1}{\text { wavelength }(\lambda)}$

$=\frac{1}{580 \times 10^{-9}}=1.72 \times 10^{6} \mathrm{~m}^{-1}$

Question 6. Find energy of each of the photons which

(I) correspond to light of frequency $3 \times 10^{15} \mathrm{~Hz}$.

(II) Have a wavelength of $0.50$ angstrom.

Solution: (I) Given: light of frequency $=3 \times 10^{15} \mathrm{~Hz}$.

The formula used: Energy (E) of one photon $=$ hy

Where $\mathrm{h}=$ Planck’s constant $=6.62 \times 10^{-34} \mathrm{Js}$

$v=$ frequency of light $=3 \times 10^{15} \mathrm{~Hz}$

Substituting the above values in the given formula of Energy:

$\mathrm{E}=\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{15} \mathrm{~Hz}\right)$

$\mathrm{E}=1.99 \times 10^{-18} \mathrm{~J}$

(II) Given: wavelength $=0.50$ angstrom

Energy (E) of a photon having wavelength $(\lambda)$ is given by the expression,

Formula: $\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$

$\mathrm{h}=$ Planck’s constant $=6.62 \times 10^{-34} \mathrm{Js}$

$\mathrm{c}=$ velocity of light in vacuum $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

we know: 1 angstrom $=10^{-10}$ meter

So $0.50$ angstrom $=0.50 \times 10^{-10}$ meter

Substituting the values in the given expression of Energy :

$\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{0.50 \times 10^{-10}}=3.97 \times 10^{-15} \mathrm{~J}$

$\therefore \mathrm{E}=3.97 \times 10^{-15} \mathrm{~J}$

Question 7. Calculate the wavelength, frequency and wave number of a light wave whose time period is $2.0 \times 10^{-10} \mathrm{~S}$

Solution: Given: time period of light $=2.0 \times 10^{-10} \mathrm{~s}$

The formula used: Frequency of light $(v)=\frac{1}{\text { time Period }}$

$=\frac{1}{2.0 \times 10^{-10} \mathrm{~s}}$

$=5.0 \times 10^{9} \mathrm{~s}^{-1}$

The formula used : Wavelength of light $(\lambda)$

$=\frac{\text { the velocity of light in a vacuum(c) }}{\text { Frequency of light (v) }}$

Where,

$\mathrm{c}=$ velocity of light in vacuum $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Substituting the value in the given expression of $\lambda$ :

$\lambda=\frac{3 \times 10^{8} \mathrm{~m} / \mathrm{s}}{5.0 \times 10^{9} \mathrm{~s}^{-1}}=6.0 \times 10^{-2} \mathrm{~m}$

Wave number $(\bar{v})$ of light $=\frac{1}{\text { wave length }(\lambda)}$

$=\frac{1}{6.0 \times 10^{-2} m}=1.66 \times 10 \mathrm{~m}^{-1}=$

$16.66 \mathrm{~m}^{-1}$

Question 8. What is the quantity of photons of light with a wavelength of $4000 \mathrm{pm}$ that gives $1 \mathrm{~J}$ of energy?

Solution: Given: wavelength of light $=4000 \mathrm{pm}$

energy $=1 \mathrm{~J}$

Formula: Energy (E) of one photon

$=$ Planck’s constant (h) $\times$ frequency of light (v)

frequency of light (v) $=\frac{\text { velocity of light }(c)}{\text { wave length }(\lambda)}$

Energy $\left(\mathrm{E}_{n}\right)$ of ‘ $n$ ‘ photons $=n h v \Rightarrow n=\frac{E_{n} \lambda}{h c}$

Where the wavelength of light $(\lambda)=4000 \mathrm{pm}=4000 \times 10^{-12} \mathrm{~m}$

The velocity of light in a vacuum $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Planck’s constant $(h)=6.62 \times 10^{-34} \mathrm{Js}$

Substituting the values in the given expression of $\mathrm{n}$ :

$\mathrm{n}=\frac{1 \times\left(4000 \times 10^{-12}\right)}{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=2.01 \times 10^{16}$

Hence, the number of photons with a wavelength of $4000 \mathrm{pm}$ and energy of $1 \mathrm{~J}$ is $2.01 \times 10^{16}$

Question 9. A photon of wavelength $4 \times 10^{-7} \mathrm{~m}$ strikes on metal surface, the work function of the metal being $2.13 \mathrm{eV}$. Calculate

(I) the energy of the photon (eV),

(II) the kinetic energy of the emission, and

(III) the velocity of the photoelectron $\left(1 \mathrm{eV}=1.6020 \times 10^{-19} \mathrm{~J}\right)$

Solution: Given: wavelength of photon $=4 \times 10^{-7} \mathrm{~m}$

work function of the metal $=2.13 \mathrm{eV}$

The energy of one photon

Formula:(E) $=$ Planck’s constant $(\mathrm{h}) \times$ frequencyof light $(v)$

$=\frac{\text { Planck’s constant }(\mathrm{h}) \times \text { velocity of light (c) }}{\text { wave length of light }(\lambda)}$

Where Planck’s constant $(h)=6.62 \times 10^{-34} \mathrm{Js}$

the velocity of light in a vacuum $(c)=3 \times 10^{8} \mathrm{~m} / \mathrm{s} \lambda=$ the wavelength of a photon $(\lambda)=4 \times 10^{-7} \mathrm{~m} / \mathrm{s}$

Substituting the values in the given expression of $E$ :

$\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{4 \times 10^{-7}}=4.965 \times 10^{-19} \mathrm{~J}$

$\mathrm{Or}=\left(\frac{4.965 \times 10^{-19}}{1.6020 \times 10^{-19}}\right) \mathrm{eV}=3.10 \mathrm{eV}$

Hence, the energy of the photon is $4.97 \times 10^{-19} \mathrm{~J}$ or $3.1020 \mathrm{eV}$

(II) Formula: The kinetic energy of emission $\mathrm{E}_{\mathrm{k}}=\mathrm{hv}-\mathrm{hv}_{0}$

$=(\mathrm{E}-\mathrm{W}) \mathrm{eV}$

$=(3.1020-2.13) \mathrm{eV}$

$=0.9720 \mathrm{eV}$

Hence, the kinetic energy of emission $\left(h v-h v_{0}\right)=0.97 \mathrm{eV}$

$=0.97 \mathrm{eV} \times 1.6020 \times 10^{-19} \mathrm{~J}$

$=1.55 \times 10^{-19} \mathrm{~J}$

(III) Formula: The velocity of a photoelectron (v) can be calculated by the expression,

$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{hv}-\mathrm{h} v_{0}$

$\Rightarrow v=\sqrt{\frac{2\left(h v-h v_{0}\right)}{m}}$

Where $\left(h v-h v_{0}\right)$ is the kinetic energy of emission in Joules = $1.55 \times 10^{-19} \mathrm{~J}$

$’ \mathrm{~m}$ ‘ is the mass of the photoelectron $=9.10939 \times 10^{-31} \mathrm{~kg}$

Substituting the values in the given expression of $\mathrm{v}$ :

$v=\frac{\sqrt{2 \times 1.55 \times 10^{-19} \mathrm{~J}}}{9.10939 \times 10^{-31} \mathrm{~kg}}$

$=\sqrt{0.3418 \times 10^{12} \mathrm{~m}^{2} \mathrm{~s}^{2}}$

$\Rightarrow \mathrm{v}=5.84 \times 10^{5} \mathrm{~ms}^{-1}$

Hence, the velocity of the photoelectron is $5.84 \times 10^{5} \mathrm{~ms}^{-1}$

Question 10. Electromagnetic radiation of wavelength $242 \mathrm{~nm}$ is just sufficient to ionise the sodium atom. Calculate the ionization energy of sodium in $\mathrm{kJ} \mathrm{mol}^{-1}$.

Solution: Given: wavelength of Electromagnetic radiation $(\lambda)=242 \mathrm{~nm}=242 \times 10^{-9} \mathrm{~m}$

Formula used: Energy of sodium atom $(\mathrm{E})=\frac{\mathrm{N}_{\mathrm{A} \mathrm{hc}}}{\lambda}$

Where, Avogadro’s number $\left(\mathrm{N}_{\mathrm{A}}\right)=6.023 \times 10^{23}$

Planck’s constant $(h)=6.62 \times 10^{-34} \mathrm{Js}$

the velocity of light in vacuum $(c)=3 \times 10^{8}$ meter $/ \mathrm{sec}$

$\mathrm{E}=\frac{\left(6.023 \times 10^{23}\right)\left(6.62 \times 10^{-34}\right) \mathrm{Js}\left(3 \times 10^{8}\right) \mathrm{ms}^{-1}}{242 \times 10^{-9} \mathrm{~m}}$

$=4.947 \times 10^{5} \mathrm{Jmol}^{-1}$

$=494.7 \times 10^{3} \mathrm{Jmol}^{-1}$

$=494 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

Question 11. A 25 -watt bulb discharges monochromatic yellow light of a wavelength of $0.57 \mu \mathrm{m}$. Ascertain the rate of discharge of quanta every second.

Solution: Given:

Power of bulb, $P=25$ Watt $=25 \mathrm{Js}^{-1}$

Wavelength $(\lambda)=0.57 \mu \mathrm{m}=0.57 \times 10^{-6}$ meter

Formula: Energy of one photon, $\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$

Substituting the values in the given expression of E:

$\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(0.57 \times 10^{-6}\right)}=34.87 \times 10^{-20} \mathrm{~J}$

Rate of emission of quanta per second

$=\frac{\text { power }}{\text { Energy }}=\frac{25}{34.87 \times 10^{-20}}=7.17 \times 10^{19} \mathrm{~s}^{-1}$

Question 12. Electrons are discharged with zero speed from a metal surface when it is presented to radiation of wavelength $6800 A$. Figure limit recurrence $\left(v_{0}\right)$ and work $\left(\mathrm{W}_{0}\right)$ of the metal.

Solution: Given: Threshold wavelength of radiation $\left(\lambda_{0}\right)=6800 A =6800 \times$ $10^{-10} \mathrm{~m}=6.8 \times 10^{-7} \mathrm{~m}$

Formula: Threshold frequency $\left(v_{0}\right)$ of the metal $=\frac{c}{\lambda_{0}}$

$=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6.8 \times 10^{-7} \mathrm{~m}}=4.41 \times 10^{14} \mathrm{~s}^{-1}$

Thus, the threshold frequency $\left(v_{0}\right)$ of the metal $=\mathrm{hv}_{0}$

$=\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(4.41 \times 10^{14} \mathrm{~s}^{-1}\right)$

$=2.922 \times 10^{-19} \mathrm{~J}$

Question 13. What is the wavelength of light transmitted when the electron in a hydrogen atom experiences move from an energy level with $\mathrm{n}=4$ to an energy level with $\mathrm{n}=2 ?$

Solution: Formula: The $n_{i}=4$ to $n_{f}=2$ transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

$\mathrm{E}=2.18 \times 10^{-18}\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^{2}}-\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}\right] \mathrm{J}$

Substituting the values in the given expression of $E$ :

$\mathrm{E}=2.18 \times 10^{-18}\left[\frac{1}{4^{2}}-\frac{1}{2^{2}}\right] J$

$\mathrm{E}=2.18 \times 10^{-18}\left[\frac{1}{16}-\frac{1}{4}\right] J$

$=2.18 \times 10^{-18}\left[\frac{1-4}{16}\right] \mathrm{J}$

$=2.18 \times 10^{-18} \times\left(-\frac{3}{16}\right) \mathrm{J}$

$\mathrm{E}=-\left(4.09 \times 10^{-19} \mathrm{~J}\right)$

formula used: $E=\frac{h c}{\text { wave length }(\lambda)}$

$\therefore$ Wavelength of light emitted $(\lambda)=\frac{\mathrm{hc}}{\mathrm{E}}$

Where, Avogadro’s number $\left(\mathrm{N}_{\mathrm{A}}\right)=6.023 \times 10^{23}$

Planck’s constant $(h)=6.62 \times 10^{-34} \mathrm{Js}$

the velocity of light in vacuum $(c)=3 \times 10^{8}$ meter $/ \mathrm{sec}$

Substituting the values given above in the formula of Wavelength of light emitted

$\lambda=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(4.0875 \times 10^{-19}\right)}$

$=4.8631 \times 10^{-7} \mathrm{~m}$

$\lambda=486.31 \times 10^{-9} \mathrm{~m} \quad\left(\therefore 10^{-9} \mathrm{~m}=1 \mathrm{~nm}\right)$

$=486 \mathrm{~nm}$

Question 14. What amount of energy is required to ionize a H molecule if the electron involves $\mathrm{n}=5$ circle? Contrast your answer and the ionization enthalpy of $\mathrm{H}$ molecule (energy required to evacuate the electron from $\mathrm{n}=1$ circle).

Solution: Given: principle quantum number $=5$

atomic number of $\mathrm{H}=1$

Formula: The expression of energy is given by,

$\mathrm{E}_{\mathrm{n}}=-\frac{(13.6) \mathrm{Z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}$

$\left(\therefore 1 \mathrm{eV} \times 1.6 \times 10^{-19}=1 \mathrm{~J}\right)$

$=-\frac{\left(2.18 \times 10^{-18}\right) \mathrm{z}^{2}}{\mathrm{n}^{2}} \mathrm{~J}$

Where,

$\mathrm{Z}=$ atomic number of the atom $\mathrm{n}$

$\mathrm{n}=$ principal quantum number

For ionization from $\mathrm{n}_{1}=5$ to $\mathrm{n}_{2}=\infty$,

$\Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_{5}$

$=\left[\left(\frac{-(13.6)(1)^{2}}{(\infty)^{2}}\right)-\left(\frac{-(13.6)(1)^{2}}{(5)^{2}}\right)\right] \mathrm{eV}$

$=0.544 \mathrm{eV}$

$\left(\therefore 1 e v \times 1.6 \times 10^{-19}=1 \mathrm{~J}\right)$

$=0.544 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{~J}$

$=0.08704 \times 10^{-18} \mathrm{~J}$

$\Delta \mathrm{E}=8.70 \times 10^{-20} \mathrm{~J}$

Hence, the energy required for ionization from $\mathrm{n}=5$ to $\mathrm{n}=\infty$ is

$\Delta \mathrm{E}=8.70 \times 10^{-20} \mathrm{~J}$

Energy required for $\mathrm{n}_{1}=1$ to $\mathrm{n}_{2}=\infty$,

$\Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_{5}$

$=\left[\left(\frac{-13.6(1)^{2}}{(\infty)^{2}}\right)-\left(\frac{-13.6(1)^{2}}{(1)^{2}}\right)\right] \mathrm{eV}$

$=13.6 \mathrm{eV}$

$\left(\therefore 1 \mathrm{eV} \times 1.6 \times 10^{-19}=1 \mathrm{~J}\right)$

$=13.6 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{~J}$

$=2.18 \times 10^{-18} \mathrm{~J}$

$\Delta \mathrm{E}=2.18 \times 10^{-18} \mathrm{~J}$

Hence, less energy is required to ionize an electron in the $5^{t h}$ orbital of the hydrogen atom as compared to that in the ground state.

Question 15. When the excited electron of an $\mathrm{H}$ atom in $\mathrm{n}=6$ drops to the ground state, the following transitions are possible:

Solution: Hence, the total number of $(5+4+3+2+1) 15$ lines will be obtained in the emission spectrum.

Alternative method:

Formula : The number of spectral lines produced when an electron in the $n^{t h}$ leVel drops down to the ground state

$\frac{n(n-1)}{2}$

Given, $n=6$

Number of spectral lines $=\frac{6(6-1)}{2}=15$

Question 16. (I) Energy associated with the fifth orbit of a hydrogen atom is calculated as:

Solution: Given: hydrogen orbit $(\mathrm{n})=5$

for hydrogen atomic number $(\mathrm{Z})=1$

$\mathrm{E}_{5}=\frac{-(13.6)(\mathrm{Z})^{2}}{(\mathrm{n})^{2}} \mathrm{eV}$

$=\frac{-(13.6)}{(5)^{2}} \mathrm{eV}$

$=\frac{-(13.6)}{25} \mathrm{eV}$

$=2.72 \mathrm{eV}$

$\left(\therefore 1 \mathrm{eV} \times 1.6 \times 10^{-19}=1 \mathrm{~J}\right)$

$=2.72 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{~J}$

$=-8.72 \times 10^{-20} \mathrm{~J}$

(II) The radius of Bohr’s $n^{t h}$ orbit for atom $\left(\mathrm{r}_{\mathrm{n}}\right)=\frac{(0.0529) \mathrm{n}^{2}}{z} \mathrm{~nm}$

Solution: We know: Hydrogen atom $\mathrm{Z}=1$

Radius For $\mathrm{n}=5$

$\mathrm{r}_{\mathrm{s}}=\frac{(0.0529) \mathrm{n}^{2}}{z}$

$r_{5}=(0.0529) \times(5)^{2} \mathrm{~nm}$

$\mathrm{r}_{5}=(0.0529) \times(25) \mathrm{nm}$

$\mathrm{r}_{5}=1.3225 \mathrm{~nm} \cong 1.32 \mathrm{~nm}$

Question 17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Solution: For the Balmer series, $n_{1}=2$. Thus, the expression of wavenumber $(\bar{v})$ is given by,

$\bar{v}=\left[\frac{1}{(2)^{2}}-\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}\right]\left(1.097 \times 10^{7} \mathrm{~m}^{-1}\right)$

Formula: Wave number $(\bar{v}) \propto \frac{1}{\text { wavelength of transition }}$.

Hence, for the longest wavelength transition, Wave number $(\bar{v})$ has to be the smallest.

For $(\bar{v})$ to be minimum, $\mathrm{n}_{2}$ should be minimum.

For the Balmer series, a transition from $\mathrm{n}_{1}=2$ to $\mathrm{n}_{2}=3$ is allowed.

Hence, taking $\mathrm{n}_{2}=3$, we get:

$\bar{v}=\left(1.097 \times 10^{7}\right)\left[\frac{1}{(2)^{2}}-\frac{1}{3^{2}}\right]$

$\bar{v}=\left(1.097 \times 10^{7}\right)\left[\frac{1}{4}-\frac{1}{9}\right]$

$=\left(1.097 \times 10^{7}\right)\left[\frac{9-4}{36}\right]$

$=\left(1.097 \times 10^{7}\right)\left[\frac{5}{36}\right]$

$\bar{v}=1.52 \times 10^{6} \mathrm{~m}^{-1}$

Question 18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state?

The ground-state electron energy is $-2.18 \times 10^{-11}$ ergs.

Solution: Energy (E) of the $\mathrm{n}^{\text {th }}$ Bohr orbit of an atom is given by

$\mathrm{E}_{5}=\frac{-\left(2.18 \times 10^{-18}\right) \mathrm{Z}^{2}}{(\mathrm{n})^{2}}$

Where $\mathrm{Z}=$ atomic number of the atom

Ground state energy $=-2.18 \times 10^{-11}$ ergs

$=-2.18 \times 10^{-11} \times 10^{-7} \mathrm{~J}$

$\left(\therefore 1\right.$ erg $\left.=10^{-7} \mathrm{~J}\right)$

$=-2.18 \times 10^{-18} \mathrm{~J}$

The energy required to shift the electron from $\mathrm{n}=1$ to $\mathrm{n}=5$ is given as:

$\Delta \mathrm{E}=\mathrm{E}_{5}-\mathrm{E}_{1}$

$=\left[\left(\frac{-\left(2.18 \times 10^{-18} \mathrm{~J}\right)(1)^{2}}{(5)^{2}}\right)-\left(\frac{-2.18 \times 10^{-18}}{1}\right)\right]$

$=\left[\left(\frac{-\left(2.18 \times 10^{-18} \mathrm{~J}\right)(1)^{2}}{25}\right)-\left(\frac{-2.18 \times 10^{-18}}{1}\right)\right]$

$=\left(2.18 \times 10^{-18}\right)\left[1-\frac{1}{25}\right]$

$=\left(2.18 \times 10^{-18}\right)\left[\frac{24}{25}\right]$

$=2.09 \times 10^{-18} \mathrm{~J}$

The wavelength of emitted light $(\lambda)=\frac{h c}{E}$

$\lambda=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(2.0928 \times 10^{-18}\right)}$

$\lambda=9.50 \times 10^{-8} \mathrm{~m}$

Question 19. The electron energy in hydrogen atom is given by $\mathrm{E}_{\mathrm{n}}=\frac{-2.18 \times 10^{-18}}{\mathrm{n}^{2}} \mathrm{~J}$ Calculate the energy required to remove an electron completely from the $\mathrm{n}=2$ orbit. What is the longest wavelength of light in $\mathrm{cm}$ that can be used to cause this transition?

Solution: $\mathrm{E}_{\mathrm{n}}=\frac{-\left(2.18 \times 10^{-18}\right)}{(\mathrm{n})^{2}} \mathrm{~J}$

The energy required for ionization from $\mathrm{n}=2$ is given by,

$\Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_{2}$

$=\left[\left(\frac{-2.18 \times 10^{-18}}{(\infty)^{2}}\right)-\left(\frac{\left(-2.18 \times 10^{-18}\right)}{(2)^{2}}\right)\right] \mathrm{J}$

$=\left[\frac{2.18 \times 10^{-18}}{4}-0\right] \mathrm{J}$

$\Delta \mathrm{E}=5.45 \times 10^{-19} \mathrm{~J}$

$\lambda=\frac{h c}{\Delta E}$

plank constant $(\mathrm{h})=6.62 \times 10^{-34}$

velocity of light $(\mathrm{c})=3 \times 10^{8}$

Here, $\lambda$ is the longest wavelength causing the transition.

$\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(0.57 \times 10^{-6}\right)}=34.87 \times 10^{-20} \mathrm{~J}$

$\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(5.45 \times 10^{-19}\right)}=3.647 \times 10^{-7} \mathrm{~m}$

$=3647 \times 10^{-10}$ meter

$\left(\therefore 10^{-10}\right.$ meter $= 1$Å)

$=3647 Å$

Question20. Calculate the wavelength of an electron moving with a velocity of $2.05 \times$ $10^{7} \mathrm{~ms}^{-1}$

Solution: Given: A mass of the particle $(\mathrm{m})=9.10939 \times 10^{-31} \mathrm{~kg}$

velocity of particle $(\mathrm{v})=2.05 \times 10^{7} \mathrm{~ms}^{-1}$

To find : wavelength of moving the particle $(\lambda)=?$

Formula used : According to de Broglie’s equation,

wave length $(\lambda)=\frac{h}{p}=\frac{h}{m v}$

Planck’s constant $(\mathrm{h})=\left(6.62 \times 10^{-34}\right) \mathrm{Js}$

Substituting the above values in the expression of $\lambda:$

$\lambda=\frac{\left(6.62 \times 10^{-34}\right) \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(2.05 \times 10^{7} \mathrm{~ms}^{-1}\right)}$

$=3.55 \times 10^{-11} \mathrm{~m}$

Question 21. The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its K.E. is $3.0 \times 10^{-25} \mathrm{~J}$, calculate its wavelength.

Solution: Given: The mass of an electron $=9.1 \times 10^{-31} \mathrm{~kg}$

K.E. of electron $=3.0 \times 10^{-25} \mathrm{~J}$

From de Broglie’s equation,

wave length $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$

$h=\left(6.62 \times 10^{-34}\right) \mathrm{Js}$

Given, Kinetic energy $(\mathrm{K} . \mathrm{E})$ of the electron $=3.0 \times 10^{-25} \mathrm{~J}$

Since $\mathrm{K} . \mathrm{E} .=\frac{1}{2} \mathrm{~m} v^{2}$

$\therefore \operatorname{Velocity}(\mathrm{v})=\frac{\frac{\sqrt{2 \mathrm{~K} E}}{\mathrm{~m}}}{1}$

$=\sqrt{\frac{2 \times\left(3.0 \times 10^{-25} \mathrm{~J}\right)}{9.10939 \times 10^{-31} \mathrm{~kg}}}$

$=\sqrt{6.5866 \times 10^{4}}$

$\mathrm{v}=811.58 \mathrm{~ms}^{-1}$

Substituting all values in the expression of $\lambda$ :

$\lambda=\frac{\left(6.62 \times 10^{-34}\right) \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(811.579 \mathrm{~ms}^{-1}\right)}$

$=8.9544 \times 10^{-7} \mathrm{~m} \cong 8.95 \times 10^{-7} \mathrm{~m}$

Question 22. (I) Write the electronic configurations of the following ions:

(a) $\mathrm{H}^{-}$

(b) $\mathrm{Na}^{+}$

(c) $\mathrm{O}_{2}^{-}$

(d) $\mathrm{F}^{-}$

(II) What are the atomic numbers of elements whose outermost electrons are represented by

(a) $3 \mathrm{~s}^{1}$

(b) $2 \mathrm{p}^{3}$ and

(c) $3 \mathrm{p}^{5}$

(III) Which atoms are indicated by the following configurations?

(a) [He] $2 \mathrm{~s}^{1}$

(b) $[\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}$

(c) $\quad[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{1}$

Solution: (I) (a) $\mathrm{H}^{-}$ ion :

The atomic number of $\mathrm{H}$ – atom is

The electronic configuration of the $\mathrm{H}$ atom is $1 \mathrm{~s}^{1}$.

A negative charge on the species indicates the addition of an electron by it.

$\therefore$ Electronic configuration of $\mathrm{H}^{-}=1 \mathrm{~s}^{2}$

(b) $\mathrm{Na}^{+}$ ion :

The atomic number of Na-atom is 11

The electronic configuration of the Na atom is $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{1}$.

A positive charge on the species indicates the loss of an electron by it.

$\therefore$ Electronic configuration of $\mathrm{Na}^{+}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{0}$ or

$1 s^{2} 2 s^{2} 2 p^{6}$

(c) $\mathrm{O}^{2-}$ ion:

The atomic number of $\mathrm{O}-$ atom is 8

The electronic configuration of $\mathrm{O}$ atom is $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}$.

A dinegative charge on the species indicates that two electrons are gained by it. $\therefore$ Electronic configuration of $0^{2-}$ ion $=1 s^{2} 2 s^{2} 2 p^{6}$

(d) $\mathrm{F}^{-}$ ion:

The atomic number of $\mathrm{F}$ – atom is 9

The electronic configuration of the $\mathrm{F}$ atom is $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$.

A negative charge on the species indicates the gain of an electron by it.

$\therefore$ Electron configuration of $\mathrm{F}^{-}$ ion $=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}$

(II) the electron configuration of the element as $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{2} \ldots \ldots$

(a) for $3 \mathrm{~s}^{1}$ the inner orbital 1 s2s $2 \mathrm{p}$ should be fulfilled

So the electron configuration of the element $=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{1}$

Number of electrons present in the atom of the element $=2+2+$ $6+1=11$

$\therefore$ The atomic number of the element $=11$

(b) for $2 \mathrm{p}^{3}$ the inner orbital $1 \mathrm{~s} 2 \mathrm{~s}$ should be fulfilled

Completing the electron configuration of the element as $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$

Number of electrons present in the atom of the element $=2+2+$ $3=7$

$\therefore$ The atomic number of the element $=7$

(c) $2 \mathrm{p}^{5}$ the inner orbital $1 \mathrm{~s}, 2 \mathrm{~s}$ should be fulfilled

Completing the electron configuration of the element as $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$

Number of electrons present in the atom of the element $=2+2+$ $5=9$

$\therefore$ Atomic number of the element $=9$

(III) (a) $[\mathrm{He}] 2 \mathrm{~s}^{1}$

The electronic configuration of the element $[\mathrm{He}]=1 \mathrm{~s}^{2}$

The electronic configuration of the element is $[\mathrm{He}] 2 \mathrm{~s}^{1}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{1}$.

$\therefore$ Atomic number of the element $=3$

Hence, the element with the electronic configuration $[\mathrm{He}] 2 \mathrm{~s}^{1}$ is lithium (Li).

(b) $[\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}$

The electronic configuration of the element

$[\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}$

The electronic configuration of the element is $[\mathrm{Ne}]=$ $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}$

$\therefore$ Atomic number of the element $=15$

Hence, the element with the electronic configuration $[\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}$ is phosphorus (P).

(c) $[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{1}$

The electronic configuration of the element

$[\mathrm{Ar}]=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}$

The electronic configuration of the element is $[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{1}=$ $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{1}$

$\therefore$ Atomic number of the element $=21$

Hence, the element with the electronic configuration $[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{1}$ is scandium $(\mathrm{Sc})$.

Question 23. What is the most reduced estimation of $n$ that permits $g$ orbitals to exist?

Solution: For g-orbitals, $l=4$.

Azimuthal quantum number (l) value $=0$ to $(\mathrm{n}-1)$

“n” of principal quantum number,

$\therefore$ For $\mathrm{l}=4$, least value of $\mathrm{n}=5$

Question 24. An electron is in one of the $3 \mathrm{~d}$ orbitals. Give the possible values of $\mathrm{n}, \mathrm{l}$ and $\mathrm{ml}$ for this electron.

Solution: For the $3 \mathrm{~d}$ orbital:

Principal quantum number $(\mathrm{n})=3$

Azimuthal quantum number (l) $=(\mathrm{n}-1)=3-1=2$

Magnetic quantum number $\left(\mathrm{m}_{1}\right)=+\mathrm{l}$ to $-\mathrm{l}=-2,-1,0,1,2$

Question 25. An atom of an element contains 29 electrons and 35 neutrons. Deduce

(i) The number of protons and

(ii) The electronic configuration of the element.

Solution: (i) For an atom to be neutral, the number of protons is equal to the number of electrons.

$\therefore$ Number of protons in the atom of the given element $=29$

(ii) Number of protons in the atom of the given element $=29$

So The electronic configuration of the atom is $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{10}$.

Question 26. Give the number of electrons in the species, $\mathrm{H}_{2}^{+}$ and $\mathrm{H}_{2}$ and $\mathrm{O}_{2}^{+}$

Solution: Atomic number of $\mathrm{H}$ -atom $=1$

In hydrogen atom : Number of proton $=$ Number of electrons $=1$

Number of electrons present in hydrogen molecule $\left(H_{2}\right)=1+1=2$

$\mathrm{H}_{2} \rightarrow \mathrm{H}_{2}^{+}+e^{-}$

$\therefore$ Number of electrons in $\left(\mathrm{H}_{2}^{+}\right)$ molecule $=2-1=1$

$\mathrm{O}_{2}^{+}:$ Atomic number of $\mathrm{O}$ -atom $=8$

In hydrogen atom : Number of proton $=$ Number of electrons $=8$

Number of electrons present in oxygen molecule $\left(\mathrm{O}_{2}\right)=8+8=16$

$\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}+e^{-}$

$\therefore$ Number of electrons in $\mathrm{O}_{2}^{+}=16-1=15$

Question 27. (I) An atomic orbital has $\mathrm{n}=3$. What are the possible values of $\mathrm{l}$ and $\mathrm{m}_{1}$ ?

(II) List the quantum numbers $\left(\mathrm{m}_{1}\right.$ and $\mathrm{l}$ ) of electrons for $3 \mathrm{~d}$ orbital.

(iii) Which of the following orbitals are possible? $1 \mathrm{p}, 2 \mathrm{~s}, 2 \mathrm{p}$ and $3 \mathrm{f}$

Solution: (I) $\quad \mathrm{n}=3$ (Given)

For a given value of $\mathrm{n}, \mathrm{l}$ can have values from 0 to $(\mathrm{n}-1)$.

$\therefore$ For $\mathrm{n}=3$

l value $=0$ to $(\mathrm{n}-1)=0,1,2$

For a given value of $1, \mathrm{ml}$ can have $(2 \mathrm{l}+1)$ values.

for $1=0$ value, $m_{0}=(2 l+1)=(2 x 0=1)$

$=1$ value $=$ so $\mathrm{m}=0$

for $\mathrm{l}=1$ value, $\mathrm{m}_{1}=(2 \mathrm{l}+1)=(2 \mathrm{x} 1=1)$

$=3$ value $=$ so $\mathrm{m}=-1,0,1$

for $1=2$ value, $m_{2}=(2 l+1)=(2 x 2=1)=5$ value

so, $\mathrm{m}=-2,-1,0,1,2$

(II) For 3 d orbital,

1 value $=0$ to $(\mathrm{n}-1)$

$\mathrm{n}=3$ so $\mathrm{l}=2$

For a given value of $1, \mathrm{~m}$ value $=(2 \mathrm{l}+1)=(2 \mathrm{x} 2+1)=5$

For $\mathrm{l}=2, \mathrm{~m}=-2,-1,0,1,2$

(III) Among the given orbitals only $2 \mathrm{~s}$ and $2 \mathrm{p}$ are possible. $1 \mathrm{p}$ and $3 \mathrm{f}$ cannot exist. For $p$ -orbital, $l=1$.

For a given value of $n, l$ can have values from 0 to $(\mathrm{n}-1)$.

For $\mathrm{l}=1$, the minimum value of $\mathrm{n}=2$

Similarly,

For f-orbital, $\mathrm{l}=4$

For $\mathrm{l}=4$, the minimum value of $\mathrm{n}=5$

$\mathrm{P}$ orbital should be started with $\mathrm{n}=2$ and $\mathrm{f}$ orbital should be started with $\mathrm{n}=5$ Hence, $1 \mathrm{p}$ and $3 \mathrm{f}$ do not exist.

Question 28. Using s, p, d notations, describe the orbital with the following quantum numbers.

(a) $\mathrm{n}=1, \mathrm{l}=0$

(b) $\quad \mathrm{n}=3 ; 1=1$;

(c) $\quad \mathrm{n}=4 ; \mathrm{l}=2$;

(d) $\quad \mathrm{n}=4 ; 1=3$;

Solution: (a) $\mathrm{n}=1, \mathrm{l}=0$ is for s orbital hence The orbital is $1 \mathrm{~s}$.

(b) For $\mathrm{n}=3$ and $\mathrm{l}=1$ for $\mathrm{p}$ orbital hence The orbital is $3 \mathrm{p}$.

(c) For $\mathrm{n}=4$ and $\mathrm{l}=2$ for $\mathrm{d}$ orbital hence The orbital is $4 \mathrm{~d}$.

(d) For $\mathrm{n}=4$ and $\mathrm{l}=3$ for $\mathrm{f}$ orbital hence The orbital is $4 \mathrm{f}$.

Question 29. Explain, giving reasons, which of the following sets of quantum numbers are not possible.

(a) $\mathrm{n}=0, \mathrm{l}=0, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

(b) $\quad \mathrm{n}=1, \mathrm{l}=0, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=-\frac{1}{2}$

(c) $\quad \mathrm{n}=1, \mathrm{l}=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

(d) $\mathrm{n}=2, \mathrm{l}=0, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=-\frac{1}{2}$

(e) $\quad \mathrm{n}=3, \mathrm{l}=3, \mathrm{~m}_{1}=-3, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

(f) $\quad \mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

Solution: (a) $\mathrm{n}=0, \mathrm{l}=0, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero. $\mathrm{n}=$

$1,1=0, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=-\frac{1}{2} ;$ it is 1s orbital

The given set of quantum numbers is possible.

(b) $\mathrm{n}=1,1=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

For the principal quantum number (n) the value of $1=(\mathrm{n}-1)=(1-1)=0$

The given set of quantum numbers is not possible because of the value of the

azimuthal quantum number (l) cannot be equal to the principal quantum number (n)

(d) $\mathrm{n}=2, \mathrm{l}=0, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=-\frac{1}{2}$

$\mathrm{n}=2, \mathrm{l}=0$ to $(\mathrm{n}-1)$ i.e. 0,1, value of $\mathrm{m}$

$=(1-1): \mathrm{m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2} \mathrm{The}$

given set of quantum numbers is not possible because the value of $\mathrm{m}$ cannot be 1

(e) $\mathrm{n}=3, \mathrm{l}=3, \mathrm{~m}_{1}=-3, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$

The given set of quantum numbers is possible. The given set of quantum numbers is not possible. For $n=3, l=0$ to $(3-1) l=0$ to 2 i.e., $0,1,2$

(f) $\mathrm{n}=3,1=0, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}$ The given set of quantum numbers is

possible. because the value of $\mathrm{m}$ cannot be 1

Question 30. How many electrons in an atom may have the following quantum numbers?

(a) $\mathrm{n}=4, \mathrm{~m}_{\mathrm{s}}=-\frac{1}{2}$

(b) $\quad \mathrm{n}=3, \mathrm{l}=0$

Solution: (a) Formula: Total number of electrons in an atom for a value of $\mathrm{n}=2 \mathrm{n}^{2}$

$\therefore$ For $\mathrm{n}=4$

Total number of electrons $=2(4)^{2}=32$

The given element has a fully filled orbital as

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$

Hence, all the electrons are paired.

$\therefore$ Number of electrons (having $\mathrm{n}=4$ and $\left.\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}\right)=\frac{32}{2}=16$

(b) $\mathrm{n}=3, \mathrm{l}=0$ indicates that the electrons are present in the $3 \mathrm{~s}$ orbital and $3 \mathrm{~s}$ orbital have only 2 electrons. Therefore, the number of electrons having $\mathrm{n}=3$ and $\mathrm{l}=0$ is 2

Question 31. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Solution: Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

Formula: $\mathrm{mvr}=\mathrm{n} \frac{\mathrm{h}}{2 \pi} \ldots .(1)$

Where, $\mathrm{n}=1,2,3, \ldots$

According to de Broglie’s equation:

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad \frac{\mathrm{hr}}{\lambda}=\mathrm{n} \frac{\mathrm{h}}{2 \pi}$

Or

$\mathrm{mv}=\frac{\mathrm{h}}{\lambda} \ldots . .(2)$ or $2 \pi \mathrm{r}=\mathrm{n} \lambda \ldots(3)$

Since ‘ $2 \pi \mathrm{r}$ ‘ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Question 32. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $\mathrm{n}=4$ to $\mathrm{n}=2$ of $\mathrm{He}^{+}$ spectrum?

Solution: For $\mathrm{He}^{+}$ ion, the wave number associated with the Balmer transition, $\mathrm{n}=4$ to $\mathrm{n}=2$ is given by:

$\bar{v}=\frac{1}{\lambda}=\mathrm{RZ}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$

Where, $\mathrm{n}_{1}=2 \mathrm{n}_{2}=4$

the atomic number of helium $(\mathrm{Z})=2$

$\bar{v}=\frac{1}{\lambda}=\mathrm{R}(2)^{2}\left(\frac{1}{4}-\frac{1}{16}\right)=4 \mathrm{R}\left(\frac{4-1}{16}\right)$

$\bar{v}=\frac{1}{\lambda}=\frac{3 R}{4}$

$\Rightarrow \lambda=\frac{4}{3 R}$

According to the question, the desired transition for hydrogen will have the same wavelength as that of $\mathrm{He}^{+}$.

$\bar{v}=\mathrm{R}(1)^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\frac{3 \mathrm{R}}{4}$

$\Rightarrow \mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\frac{3 \mathrm{R}}{4} \ldots(1)$

By hit and trial method, the equality is given by equation (1) is true only when $\mathrm{n}_{1}=1$ and $\mathrm{n}_{2}=2$

The transition for $\mathrm{n}_{2}=2$ to $\mathrm{n}=1$ in hydrogen spectrum would have the same wavelength as Balmer transition $\mathrm{n}=4$ to $\mathrm{n}=2$ of $\mathrm{He}^{+}$ spectrum.

Question 33. Calculate the energy required for the process

$\mathrm{He}_{(\mathrm{g})}^{+} \rightarrow \mathrm{He}_{(\mathrm{g})}^{2+}+\mathrm{e}^{-}$

The ionization energy for the $\mathrm{H}$ atom in the ground state is $2.18 \times 10^{-18} \mathrm{~J}$ atom $^{-1}$

Solution: Energy associated with hydrogen-like species is given by,

$\mathrm{E}_{\mathrm{n}}=-2.18 \times 10^{-18}\left(\frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}\right) \mathrm{J}$

For ground state of hydrogen atom,

$\Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_{1}$

$=0-\left[-\left(2.18 \times 10^{-18} \frac{(1)^{2}}{(1)^{2}}\right] \mathrm{J}\right.$

$\Delta \mathrm{E}=2.18 \times 10^{-18} \mathrm{~J}$

For the given process,

$\mathrm{He}_{(\mathrm{g})}^{+} \rightarrow \mathrm{He}_{(\mathrm{g})}^{2+}+\mathrm{e}^{-}$

An electron is removed from $\mathrm{n}=1$ to $\mathrm{n}=\infty$.

$\Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_{1}$

$=0-\left[-\left(2.18 \times 10^{-18} \frac{(2)^{2}}{(1)^{2}}\right] \mathrm{J}\right.$

$\Delta \mathrm{E}=8.72 \times 10^{-18} \mathrm{~J}$

$\therefore$ The energy required for the process $8.72 \times 10^{-18} \mathrm{~J}$.

Question 34. If the diameter of a carbon atom is $0.15 \mathrm{~nm}$, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length $20 \mathrm{~cm}$ long.

Solution: $1 \mathrm{~m}=100 \mathrm{~cm}$

$1 \mathrm{~cm}=10-2 \mathrm{~m}$

Length of the scale $=20 \mathrm{~cm}$

$=20 \times 10^{-2} \mathrm{~m}$

Diameter of a carbon atom $=0.15 \mathrm{~nm}$

$=0.15 \times 10^{-9} \mathrm{~m}$

One carbon atom occupies $0.15 \times 10^{-9} \mathrm{~m}$.

$\therefore$ Number of carbon atoms that can be placed in a straight line $\frac{20 \times 10^{-2} \mathrm{~m}}{0.15 \times 10^{-9} \mathrm{~m}}$

$=133.33 \times 10^{7}$

$=1.33 \times 10^{9}$

Question 35. $2 \times 10^{8} \mathrm{~m}$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is $2.4 \mathrm{~cm}$.

Solution: Length of the given arrangement $=2.4 \mathrm{~cm}$

Number of carbon atoms present $=2 \times 10^{8}$

Diameter of carbon atom

$\frac{2.4 \times 10^{-2} \mathrm{~m}}{2 \times 10^{8} \mathrm{~m}} \quad \therefore$ Radius of carbon atom $=1.2 \times 10^{-10} \mathrm{~m}$

$=\frac{\text { Diameter }}{2}$

$=\frac{1.2 \times 10^{-10} \mathrm{~m}}{2}$

$=6.0 \times 10^{-11} \mathrm{~m}$

Question 36. The diameter of zinc atom is 2.6 Å.Calculate

(a) Radius of zinc atom in pm and

(b) Number of atoms present in a length of $1.6 \mathrm{~cm}$ if the zinc atoms are arranged side by side lengthwise.

Solution: (a) Formula: Radius of carbon atom $=\frac{\text { Diameter }}{2}$

$=\frac{2.6}{2}$

$=1.3 \times 10^{-10} \mathrm{~m}$

$=130 \times 10^{-12} \mathrm{~m}$

$=130 \mathrm{pm}$

(b) Length of the arrangement $=1.6 \mathrm{~cm}$

$=1.6 \times 10^{-2} \mathrm{~m}$

Diameter of zinc atom $=1.6 \times 10^{-10} \mathrm{~m}$

$\therefore$ Number of zinc atoms present in the arrangement

$=1 \cdot \frac{6 \times 10^{-2} \mathrm{~m}}{2.6 \times 10^{-10} \mathrm{~m}}$

$=0.6153 \times 10^{8} \mathrm{~m}$

$=6.153 \times 10^{7} \mathrm{~m}$

Question 37. The diameter of zinc atom is 2.6 Å. Calculate

(a) radius of zinc atom in pm and

(b) number of atoms present in a length of $1.6 \mathrm{~cm}$ if the zinc atoms are arranged side by side lengthwise.

Solution: (a) Radius of carbon atom $=\frac{\text { Diameter }}{2}$

$=\frac{2.6}{2}$

$=1.3 \times 10^{-10} \mathrm{~m}$

$=130 \times 10^{-12} \mathrm{~m}$

$=130 \mathrm{pm}$

(b) Length of the arrangement $=1.6 \mathrm{~cm}$

$=1.6 \times 10^{-2} \mathrm{~m}$

Diameter of zinc atom $=1.6 \times 10^{-10} \mathrm{~m}$

Therefore, Number of zinc atoms present in the arrangement

$=\frac{1.6 \times 10^{-2} \mathrm{~m}}{2.6 \times 10^{-10} \mathrm{~m}}$

$=0.6153 \times 10^{8} \mathrm{~m}$

$=6.153 \times 10^{7} \mathrm{~m}$

Question 38. A certain particle carries $2.5 \times 10^{-16} \mathrm{C}$ of static electric charge. Calculate the number of electrons present in it.

Solution: Charge on one electron $=1.6022 \times 10^{-19} \mathrm{C}$

$\Rightarrow 1.6022 \times 10^{-19} \mathrm{C}$ charge is carried by 1 electron.

Therefore, Number of electrons carrying a charge of

$2.5 \times 10^{-16} \mathrm{C} \frac{1}{1.6022 \times 10^{-19} \mathrm{C}}\left(2.5 \times 10^{-16} \mathrm{C}\right)$

$=1.560 \times 10^{3} \mathrm{C}$

$=1560 \mathrm{C}$

Question 39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $-1.282 \times$ $10^{-18} \mathrm{C}$, calculate the number of electrons present on it.

Solution: Charge on the oil drop $=1.282 \times 10^{-18} \mathrm{C}$

Charge on one electron $=1.6022 \times 10^{-19} \mathrm{C}$

Therefore, Number of electrons present on the oil drop

$\frac{1.282 \times 10^{-18} \mathrm{C}}{1.6022 \times 10^{-19} \mathrm{C}}$

$=0.8001 \times 10^{1}$

$=8.0$

Question 40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the $\alpha$ -particles. If the thin foil of light atoms like Aluminium etc. is used, what difference would be observed from the above results?

Solution: A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.

Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of $\alpha$ -particles (positively charged).

Question 41. Symbols ${ }_{35}^{79} \mathrm{Br}$ and ${ }^{79} \mathrm{Br}$ can be written, whereas symbols ${ }_{79}^{35} \mathrm{Br}$ and ${ }^{35} \mathrm{Br}$ are not acceptable. Answer briefly.

Solution: The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}$.

Hence, ${ }_{35}^{79} \mathrm{Br}$ is acceptable but ${ }_{79}^{35} \mathrm{Br}$ is not acceptable.

${ }^{79} \mathrm{Br}$ can be written but ${ }^{35} \mathrm{Br}$ cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.

Question 42. An element with mass number 81 contains $31.7 \%$ more neutrons as compared to protons. Assign the atomic symbol.

Solution: Let the number of protons in the element be $\mathrm{x}$.

Therefore, Number of neutrons in the element $=\mathrm{x}+31.7 \%$ of $\mathrm{x}$

$=\mathrm{x}+0.317 \mathrm{x}$

$=1.317 \mathrm{x}$

According to the question,

Mass number of the element $=81$

Therefore, (Number of protons $+$ number of neutrons $)=81$

$\Rightarrow x+1.317 x=81 \simeq 35$

$2.317 x=81 x$

$=\frac{81}{2.317}$

$=34.95 \cong 35$

Hence, the number of protons in the element i.e., $\mathrm{x}$ is 35 .

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35 .

Therefore, the atomic symbol of the element is ${ }_{35}^{81} \mathrm{Br}$

Question 43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains $11.1 \%$ more neutrons than the electrons, find the symbol of the ion.

Solution: Let the number of electrons in the ion carrying a negative charge be $\mathrm{x}$.

$=\mathrm{x}+0.111 \mathrm{x}$

$=1.111 \mathrm{x}$

Number of electrons in the neutral atom $=(\mathrm{x}-1)$

(When an ion carries a negative charge, it carries an extra electron)

Therefore, Number of protons in the neutral atom $=\mathrm{x}-1$

Given,

Mass number of the ion $=37$

Therefore, $(\mathrm{x}-1)+1.111 \mathrm{x}=37$

$2.111 x=38 x$

$=18$

Therefore, The symbol of the ion is ${ }_{17}^{37} \mathrm{Cl}^{-}$

Question 44. An ion with mass number 56 contains 3 units of positive charge and $30.4 \%$ more neutrons than electrons. Assign the symbol to this ion.

Solution: Let the number of electrons present in ion $A^{3+}$ be $x$.

Therefore, Number of neutrons in $\mathrm{it}=x+30.4 \%$ of $x=1.304 x$

Since the ion is tripositive,

$\Rightarrow$ Number of electrons in neutral atom $=x+3$

Therefore, Number of protons in neutral atom $=x+3$

Given: Mass number of the ion $=56$

Therefore, $(x+3)(1.304 x)=56$

$2.304 x=53$

$x=532.304$

$x=23$

Therefore, Number of protons $=x+3=23+3=26$

Therefore, The symbol of the ion $\frac{56}{26} \mathrm{Fe}^{+}$

Question 45. Arrange the following type of radiations in increasing order of frequency:

(b) amber light from traffic signal

Radiation from FM radio $<$ amber light $<$ radiation from microwave oven $<\mathrm{X}-$ rays $<$ cosmic rays
Cosmic rays $Question 46. Nitrogen laser produces a radiation at a wavelength of$337.1 \mathrm{~nm}$. If the number of photons emitted is$5.6 \times 1024$, calculate the power of this laser. Solution: Power of laser$=$Energy with which it emits photons Formula: Power$=\mathrm{E}=$Nhc$\lambda$Where,$\mathrm{N}=$number of photons emitted$\mathrm{h}=$Planck’s constant$\mathrm{c}=$velocity of radiation$\lambda=$wavelength of radiation Substituting the values in the given expression of Energy (E):$E=\frac{\left(5.6 \times 10^{24}\right)(6.62 \times 10-34 \mathrm{Js})\left(3 \times 10^{8} \mathrm{~ms}-1\right)}{337.1 \times 10^{-9} \mathrm{~m}}=0.3302 \times 10^{7} \mathrm{~J}=3.33 \times 10^{6} \mathrm{~J}$Hence, the power of the laser is$3.33 \times 10^{6} \mathrm{~J}$Question 47. Neon gas is generally used in the sign boards. If it emits strongly at$616 \mathrm{~nm}$, calculate (a) the frequency of emission, (b) distance traveled by this radiation in$30 \mathrm{~s}$(c) energy of quantum and (d) number of quanta present if it produces$2 \mathrm{~J}$of energy. Solution: Wavelength of radiation emitted$=616 \mathrm{~nm}=616 \times 10^{-9} \mathrm{~m}$(Given) (a) Frequency of emission (v)$v=c \lambda$Where,$\mathrm{c}=$velocity of radiation$\lambda=$wavelength of radiation Substituting the values in the given expression of$(v)$:$v=3 \times \frac{108 \mathrm{~m}}{\mathrm{~s} 616} \times 10^{-9} \mathrm{~m}=4.87 \times 10^{8} \times 10^{9} \times 10^{-3} \mathrm{~s}^{-1}\mathrm{v}=4.87 \times 10^{14} \mathrm{~s}^{-1}$Frequency of emission$(v)=4.87 \times 10^{14} \mathrm{~s}^{-1}$(b) We know: Velocity of radiation,$\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$Distance travelled by this radiation in$30 \mathrm{~s}=\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)(30 \mathrm{~s})=9 \times 10^{9} \mathrm{~m}$(c) Energy of quantum$(\mathrm{E})=\mathrm{h} v=\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(4.87 \times 10^{14} \mathrm{~s}^{-1}\right)$Energy of quantum$(\mathrm{E})=32.27 \times 10^{-20} \mathrm{~J}$Therefore,$32.27 \times 10^{-20} \mathrm{~J}$of energy is present in 1 quantum. Number of quanta in$2 \mathrm{~J}$of energy$\frac{2 \mathrm{~J}}{32.27 \times 10^{-20} \mathrm{~J}}=6.19 \times 10^{18}=6.2 \times 10^{18}$Question 48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of$3.15 \times 10^{-18} \mathrm{~J}$from the radiations of$600 \mathrm{~nm}$, calculate the number of photons received by the detector. Solution: We know: velocity of radiation (c)$=3 \times 10^{8} \mathrm{~ms}^{-1}$Planck’s constant$(\mathrm{h})=6.62 \times 10^{-34} \mathrm{Js}$Given: Wavelength$(\lambda)=600 \mathrm{~nm}$Energy received by the photons$=3.15 \times 10^{-18} \mathrm{~J}$From the expression of energy of one photon (E), Formula:$\mathrm{E}=$hc$\lambda$Where,$\lambda=$wavelength of radiation$\mathrm{h}=$Planck’s constant$\mathrm{c}=$velocity of radiation Substituting the values in the given expression of$\mathrm{E}$:$E=\frac{\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{600 \times 10^{-9}}=3.313 \times 10^{-19} \mathrm{~J}$Energy of one photon$=3.313 \times 10^{-19} \mathrm{~J}$Number of photons received with$3.15 \times 10^{-18} \mathrm{~J}$of energy$=\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}=9.5 \approx 10$Question 49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of$2 \mathrm{~ns}$and the number of photons emitted during the pulse source is$2.5 \times 10^{15} \mathrm{~J}$, calculate the energy of the source. Solution: Frequency of radiation (v),$v=12.0 \times 10^{-9} \mathrm{~s}v=5.0 \times 10^{8} \mathrm{~s}^{-1}$Energy (e)$\quad$of source$=\mathrm{Nhv}$Where,$\mathrm{N}=$number of photons emitted$\mathrm{h}=$Planck’s constant$v=$frequency of radiation Substituting the values in the given expression of (E):$\mathrm{E}=\left(2.5 \times 10^{15}\right)\left(6.62 \times 10^{-34} \mathrm{Js}\right)(5.0 \times 108 \mathrm{~s}-1)\mathrm{E}=8.282 \times 10^{-10} \mathrm{~J}$Hence, the energy of the source (e) is$8.282 \times 10^{-10} \mathrm{~J}$. Question 50. The work function for cesium atom is$1.9 \mathrm{eV}$. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the cesium element is irradiated with a wavelength$500 \mathrm{~nm}$, calculate the kinetic energy and the velocity of the ejected photoelectron. Solution: It is given that the work function$\left(\mathrm{W}_{0}\right)$for cesium atom is$1.9 \mathrm{eV}$. (a) From the$\mathrm{W}_{0}=\mathrm{hc} \lambda_{0}$expression, we get:$\lambda_{0}=\mathrm{hcW}_{0}$Where,$\lambda_{0}=$threshold wavelength$\mathrm{h}=$Planck’s constant$c=$velocity of radiation Substituting the values in the given expression of$\left(\lambda_{0}\right):\lambda_{0}=\frac{\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(1.9 \times 1.602 \times 10^{-19} \mathrm{~J}\right)}=6.53 \times 10^{-7} \mathrm{~m}$Hence, the threshold wavelength$\left(\lambda_{0}\right)$is$653 \mathrm{~nm}$. (b) Formula: From the expression,$W_{0}=h v_{0}$, we get:$v_{0}=W_{0} h$Where,$v_{0}=$threshold frequency$\mathrm{h}=$Planck’s constant Substituting the values in the given expression of$v_{0}$:$v_{0}=\frac{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}{6.62 \times 10^{-34} \mathrm{Js}}$(i e.,$\left.V=1.602 \times 10^{-19} \mathrm{~J}\right)v_{0}=4.593 \times 10^{14} \mathrm{~s}^{-1}$Hence, the threshold frequency of radiation$\left(v_{0}\right)$is$4.593 \times 10^{14} \mathrm{~s}^{-1}$(c) According to the question: Wavelength used in irradiation$(\lambda)=500 \mathrm{~nm}$Formula: Kinetic energy$=\mathrm{h}(v-v 0)=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\left(6.62 \times 10^{-34} \mathrm{Js}\right)(3.0 \times 108 \mathrm{~ms}-1)\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right)=\left(1.9878 \times 10^{-26} \mathrm{Jm}\right)\left[\frac{(653-500) 10^{-9} \mathrm{~m}}{(653)(500) 10^{-18} \mathrm{~m}^{2}}\right]=\frac{\left(1.9878 \times 10^{-26}\right)\left(153 \times 10^{9}\right)}{(653)(500) \mathrm{J}}=9.3149 \times 10^{-20} \mathrm{~J}$Kinetic energy of the ejected photoelectron$=9.3149 \times 10^{-20} \mathrm{~J}$Since K. E =$\frac{1}{2} \mathrm{mv}^{2}=9.3149 \times 10^{-20} \mathrm{~J}v=\sqrt{\frac{2\left(9.3149 \times 10^{-20} \mathrm{~J}\right)}{9.10939 \times 10-31}}=\sqrt{2.0451 \times 10^{11} \mathrm{~m}^{2} \mathrm{~s}^{2}}=4.52 \times 10^{5} \mathrm{~ms}^{-1}$Question 51. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) Threshold wavelength and, (b) Planck’s constant. Solution: (a) Assuming the threshold wavelength to be$\lambda_{0} \mathrm{~nm}\left(=\lambda_{0} \times 10^{-} 9 \mathrm{~m}\right)$, the kinetic energy of the radiation is given as:$\mathrm{h}\left(v-v_{0}\right)=\frac{1}{2} \mathrm{mv}^{2}$Three different equalities can be formed by the given value as:$\mathrm{h}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} \mathrm{mv}^{2}h c\left(\frac{1}{500 \times 10^{9}}-\frac{1}{\lambda_{0} \times 10-9 m}\right)=\frac{1}{2} m\left(2.55 \times 10^{+5} \times 10^{-2} \mathrm{~ms}^{-1}\right)\frac{h c}{10^{-9} m}\left(\frac{1}{500}-\frac{1}{\lambda_{0}}\right)=12 m\left(2.55 \times 10^{+3} m s^{-1}\right)^{2}$(1) Similarly,$\frac{h c}{10^{-9} m}\left(\frac{1}{450}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2 m}\left(3.45 \times 10^{+3} \mathrm{~ms}^{-1}\right)^{2}$(2)$\frac{h c}{10^{-9} m}\left(\frac{1}{400}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2 m}\left(5.35 \times 10^{+3} \mathrm{~ms}^{-1}\right)^{2}$(3) Dividing equation (3) by equation (1):$\frac{\left[\frac{\lambda_{0}-400}{400 \lambda_{0}}\right]}{\left[\frac{\lambda_{0}-500}{500 \lambda_{0}}\right]}=\frac{\left(5.35 \mid \times 10^{+3} \mathrm{~ms}^{-1}\right)^{2}}{\left(2.55 \mid \times 10^{+3} \mathrm{~ms}^{-1}\right)^{2}}\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=4.4017717.6070 \lambda_{0}-5 \lambda_{0}=8803.537-2000\lambda_{0}=\frac{6805.537}{12.607}\lambda_{0}=539.8 \mathrm{~nm}\lambda_{0}=540 \mathrm{~nm}$So, threshold wavelength$\left(\lambda_{0}\right)=540 \mathrm{~nm}$(b) of the question is not done due to the incorrect values of velocity given in the question.$\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=(5.35)^{2}(2.55)^{2}=\frac{28.6225}{6.5025}\frac{5 \lambda_{0}-2000}{4 \lambda 0-2000}=4.4017717.6070 \lambda_{0}-5 \lambda_{0}=8803.537-2000\lambda_{0}=\frac{6805.537}{12.607}\lambda_{0}=539.8 \mathrm{~nm}\lambda_{0}=540$Question 52. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of$0.35 \mathrm{~V}$when the radiation$256.7 \mathrm{~nm}$is used. Calculate the work function for silver metal. Solution: Given: Stopping potential$=0.35 \mathrm{~V}$Wavelength of radiation$=256.7 \mathrm{~nm}$We know: velocity of radiation$(\mathrm{c})=3 \times 10^{8} \mathrm{~ms}^{-1}$Planck’s constant$(\mathrm{h})=6.62 \times 10^{-34} \mathrm{Js}$From the principle of conservation of energy, the energy of an incident photon (e) is equal to the sum of the work function$\left(\mathrm{W}_{0}\right)$of radiation and its kinetic energy (K.E) i.e., Fomula:$E=W_{0}+K . E\Rightarrow \mathrm{W}_{0}=\mathrm{E}-\mathrm{K} . \mathrm{E}$Formula: Energy of incident photon$(\mathrm{E})=\frac{\mathrm{hc}}{\lambda}$Where,$\mathrm{c}=$velocity of radiation$\mathrm{h}=$Planck’s constant$\lambda=$wavelength of radiation Substituting the values in the given expression of$E$:$\mathrm{E}=\frac{\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(256.7 \times 10^{-9} \mathrm{~m}\right)}=7.744 \times 10^{-19} \mathrm{~J}=\frac{7.744 \times 10^{-19}}{1.602 \times 10^{-19}} \mathrm{eV}\mathrm{E}=4.83 \mathrm{eV}$The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,$\mathrm{K} . \mathrm{E}=0.35 \mathrm{~V}\mathrm{K} . \mathrm{E}=0.35 \mathrm{eV}$Therefore, Work function,$\mathrm{W}_{0}=\mathrm{E}-\mathrm{K} . \mathrm{E} \quad[$Formula$]=4.83 \mathrm{eV}-0.35 \mathrm{eV}=4.48 \mathrm{eV}\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}5 \lambda_{0}-20004 \lambda_{0}-2000=4.4017717.6070 \lambda_{0}-5 \lambda_{0}=8803.537-2000\lambda_{0}=\frac{6805.537}{12.607}\lambda_{0}=539.8 \mathrm{~nm}\lambda_{0} \simeq 540 \mathrm{~nm}$Question 53. If the photon of the wavelength$150 \mathrm{pm}$strikes an atom and one of its inner bound electrons is ejected out with a velocity of$1.5 \times 10^{7} \mathrm{~ms}^{-1}$, calculate the energy with which it is bound to the nucleus. Solution: Given: Wavelength of photon$=150 \mathrm{pm}$Velocity of ejected electrons$=1.5 \times 10^{7} \mathrm{~ms}^{-1}$We know: velocity of radiation$(\mathrm{c})=3 \times 10^{8} \mathrm{~ms}^{-1}$Planck’s constant$(\mathrm{h})=6.62 \times 10^{-34} \mathrm{Js}$Formula: Energy of incident photon (E)is given by,$\mathrm{E}=\frac{\mathrm{hc}}{\lambda}\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(150 \times 10^{-12}\right)}=1.3252 \times 10^{-15} \mathrm{~J}\simeq 13.252 \times 10^{-16} \mathrm{~J}$Energy of the electron ejected$(\mathrm{K} . \mathrm{E})=\frac{1}{2} \mathrm{~m}_{\mathrm{e}} v^{2}=12\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(1.5 \times 10^{7} \mathrm{~ms}^{-1}\right)^{2}=10.2480 \times 10^{-17} \mathrm{~J}=1.025 \times 10^{-16} \mathrm{~J}$Hence, the energy with which the electron is bound to the nucleus can be obtained as:$=\mathrm{E}-\mathrm{K} . \mathrm{E}=13.252 \times 10^{-16} \mathrm{~J}-1.025 \times 10^{-16} \mathrm{~J}=12.227 \times 10^{-16} \mathrm{~J}=\frac{12.227 \times 10^{-16}}{1.602 \times 10^{-19}} \mathrm{eV}=7.6 \times 10^{3} \mathrm{eV}\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=4.4017717.6070 \lambda_{0}-5 \lambda_{0}=8803.537-2000\lambda_{0}=\frac{6805.537}{12.607}\lambda_{0}=539.8 \mathrm{~nm}\lambda_{0} \simeq 540 \mathrm{~nm}$Question 54. Emission transitions in the Paschen series end at orbit$\mathrm{n}=3$and start from orbit$\mathrm{n}$and can be represented as$\mathrm{v}=3.29 \times 10^{15} \mathrm{~Hz}\left[\frac{1}{3^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$Calculate the value of$\mathrm{n}$if the transition is observed at$1285 \mathrm{~nm}$. Find the region of the spectrum. Solution: Given: Wavelength of transition$=1285 \mathrm{~nm}=1285 \times 10^{-9} \mathrm{~m}v=3.29 \times 10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)$Formula: Since$v=\frac{c}{\lambda}=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{1285 \times 10-9 \mathrm{~m}}$Now,$v=2.33 \times 10^{14} \mathrm{~s}^{-1}$Substituting the value of$v$in the given expression,$3.29 \times 10^{15}\left(\frac{1}{9}-\frac{1}{\mathrm{n}^{2}}\right)=2.33 \times 10^{14}\frac{1}{9}-\frac{1}{n^{2}}=\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}}\frac{1}{9}-0.7082 \times 10^{-1}=\frac{1}{n^{2}}\Rightarrow \frac{1}{\mathrm{n}^{2}}=1.1 \times 10^{-1}-0.7082 \times 10^{-1}\frac{1}{n^{2}}=4.029 \times 10^{-2} n=14.029 \times 10^{-2}\mathrm{n}=4.98n=5$Hence, for the transition to be observed at$1285 \mathrm{~nm}, \mathrm{n}=5$. The spectrum lies in the infra-red region. Question 55. Calculate the wavelength for the emission transition if it starts from the orbit having radius$1.3225 \mathrm{~nm}$and ends at$211.6 \mathrm{pm}$. Name the series to which this transition belongs and the region of the spectrum. Solution: Radius of the initial orbit$=1.3225 \mathrm{~nm}$Radius of the final orbit$=211.6 \mathrm{pm}$Formula: The radius of the$\mathrm{n}$th orbit of hydrogen-like particles is given by,$r=\frac{0.529 n^{2}}{Z} Å\mathrm{r}=\frac{5.29 \mathrm{n}^{2}}{\mathrm{Z}} \mathrm{pm}$For radius$\left(r_{1}\right)=1.3225 \mathrm{~nm}=1.32225 \times 10^{-9} \mathrm{~m} \mathrm{n}_{1}^{2}=\frac{\mathrm{r}_{1} \mathrm{z}}{52.9}=1322.25 \times 10^{-12} \mathrm{~m} \mathrm{n}_{1}^{2}=\frac{1322.25 \mathrm{Z}}{52.9}=1322.25 \mathrm{pm}$Similarly,$\mathrm{n}_{2}^{2}=\frac{211.6 \mathrm{Z}}{52.9}\frac{\mathrm{n}_{1}^{2}}{\mathrm{n}_{2}^{2}}=\frac{1322.5}{211.6}\frac{\mathrm{n}_{1}^{2}}{\mathrm{n}_{2}^{2}}=6.25\frac{\mathrm{n}^{1}}{\mathrm{n}^{2}}=2.5\frac{\mathrm{n}^{1}}{\mathrm{n}^{2}}=\frac{25}{10}=\frac{5}{2}\Rightarrow \mathrm{n} 1=5$and$\mathrm{n} 2=2$Thus, the transition is from the 5 th orbit to the 2 nd orbit. It belongs to the Balmer series. wave number$(\bar{v})$for the transition is given by,$1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right) \mathrm{m}^{-1}=1.097 \times 10^{7} \mathrm{~m}^{-1}\left(\frac{21}{100}\right)=2.303 \times 10^{6} \mathrm{~m}^{-1}$Wavelength$(\lambda)$associated with the emission transition is given by,$\lambda=1 v=12.303 \times 10^{6} \mathrm{~m}^{-1}=0.434 \times 10^{-6} \mathrm{~m} \lambda=434 \mathrm{~nm}$Question 56. Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is$1.6 \times 10^{6} \mathrm{~ms}^{-1}$, calculate de Broglie wavelength associated with this electron. Solution: Given: velocity of the electron$=1.6 \times 10^{6} \mathrm{~ms}^{-1}$We know: Mass of electron$=9.103939 \times 10^{-31} \mathrm{~kg}$Formula: From de Broglie’s equation,$\lambda=\frac{\mathrm{h}}{\mathrm{m} v}=\frac{\left(6.62 \times 10^{-34}\right)}{9.103939 \times 10^{-31} \mathrm{~kg}\left(1.6 \times 10^{6} \mathrm{~ms}^{-1}\right)}=4.55 \times 10^{-10} \mathrm{~m} \lambda=455 \mathrm{pm}$de Broglie’s wavelength associated with the electron is$455 \mathrm{pm}$. Question 57. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is$800 \mathrm{pm}$, calculate the characteristic velocity associated with the neutron. Solution: Given: The wavelength used$=800 \mathrm{pm}$We know: Planck’s constant$=6.62 \times 10^{-34}$Mass of the particle$=1.67 \times 10^{-27} \mathrm{~kg}$From de Broglie’s equation,$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}v=\frac{h}{m \lambda}$Where,$\mathrm{v}=$velocity of particle (neutron)$\mathrm{h}=$Planck’s constant$\mathrm{m}=$mass of particle (neutron)$\lambda=$wavelength Substituting the values in the expression of velocity (v),$\mathrm{E}=\frac{\left(6.62 \times 10^{-34}\right)}{\left(1.67 \times 10^{-27}\right)\left(800 \times 10^{-12} \mathrm{~m}\right)}=4.94 \times 10^{2} \mathrm{~J}=494 \mathrm{~ms}^{-1}$Velocity associated with the neutron$=494 \mathrm{~ms}^{-1}$Question 58. If the velocity of the electron in Bohr’s first orbit is$2.19 \times 10^{6} \mathrm{~ms}^{-1}$, calculate the de Broglie wavelength associated with it. Solution: Given: velocity of the electron in first orbit$=2.19 \times 10^{6} \mathrm{~ms}^{-1}$We know: The value of planck’s constant$=6.62 \times 10^{-34}$Formula: According to de Broglie’s equation,$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$Where,$\lambda=$wavelength associated with the electron$\mathrm{h}=$Planck’s constant$\mathrm{m}=$mass of electron$\mathrm{v}=$velocity of electron Substituting the values in the expression of$\lambda$:$\lambda=\frac{h}{m v}=\frac{\left(6.62 \times 10^{-34}\right)}{9.103939 \times 10^{-31} \mathrm{~kg}\left(2.19 \times 10^{6} \mathrm{~ms}^{-1}\right)}=3.32 \times 10^{-10} \mathrm{~m} \lambda=332 \mathrm{pm}$Question 59. The velocity associated with a proton moving in a potential difference of$1000 \mathrm{~V}$is$4.37 \times 10^{5} \mathrm{~ms}^{-1}$. If the hockey ball of mass$0.1 \mathrm{~kg}$is moving with this velocity, calculate the wavelength associated with this velocity. Solution: Given: Potential difference$=1000 \mathrm{~V}$Velocity of the proton$=4.37 \times 10^{5} \mathrm{~ms}^{-1}$Mass of the hockey ball$=0.1 \mathrm{~kg}$We know: Planck’s constant$(\mathrm{h})=6.62 \times 10^{-34}$Formula: According to de Broglie’s expression,$\lambda=\frac{\mathrm{h}}{\mathrm{m} v}=\frac{\left(6.62 \times 10^{-34}\right)}{0.1 \mathrm{~kg}\left(4.37 \times 10^{5} \mathrm{~ms}^{-1}\right)}=1.516 \times 10^{-38} \mathrm{~m}$Question 60. If the position of the electron is measured within an accuracy of$+0.002 \mathrm{~nm}$, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is$\mathrm{h} / 4 \pi_{\mathrm{m}} \times 0.05 \mathrm{~nm}$, is there any problem in defining this value. Solution: Given: Accuracy in measurement of position$=+0.002 \mathrm{~nm}$Momentum of the electron$=\frac{\mathrm{h}}{4 \pi_{\mathrm{m}}} \times 0.05 \mathrm{~nm}$From Heisenberg’s uncertainty principle, Where,$\Delta \mathrm{x}=$uncertainty in position of the electron$\Delta \mathrm{p}=$uncertainty in momentum of the electron Substituting the values in the expression of$\Delta \mathrm{p}$:$=2.637 \times 10^{-23} \mathrm{Jsm}^{-1}\Delta \mathrm{p}=2.637 \times 10^{-23} \mathrm{kgms}^{-1}\left(1 \mathrm{~J}=1 \mathrm{kgms} 2 \mathrm{~s}^{-1}\right)$Uncertainty in the momentum of the electron$=2.637 \times 10^{-23} \mathrm{kgms}^{-1} .=1.055 \times 10^{-24} \mathrm{kgms}^{-1}$Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined. Question 61. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination (s) has/have the same energy lists: 1.$\mathrm{n}=4, \mathrm{l}=2, \mathrm{ml}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$2.$\mathrm{n}=3, \mathrm{l}=2, \mathrm{ml}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$3.$\mathrm{n}=4, \mathrm{l}=1, \mathrm{ml}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2$4.$\mathrm{n}=3, \mathrm{l}=2, \mathrm{ml}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$5.$\mathrm{n}=3, \mathrm{l}=1, \mathrm{ml}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$6.$\mathrm{n}=4, \mathrm{l}=1, \mathrm{ml}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2$Solution: For$\mathrm{n}=4$and$\mathrm{l}=2$, the orbital occupied is$4 \mathrm{~d}$. For$\mathrm{n}=3$and$\mathrm{l}=2$, the orbital occupied is$3 \mathrm{~d}$. For$\mathrm{n}=4$and$\mathrm{l}=1$, the orbital occupied is$4 \mathrm{p}$. Hence, the six electrons i.e.,$1,2,3,4,5$, and 6 are present in the$4 \mathrm{~d}, 3 \mathrm{~d}, 4 \mathrm{p}, 3 \mathrm{~d}, 3 \mathrm{p}$, and$4 \mathrm{p}$orbitals respectively. Therefore, the increasing order of energies is$5(3 p)<2(3 d)=4(3 \mathrm{~d})<3(4 p)=6(4 p)<1(4 d)$Question 62. The bromine atom possesses 35 electrons. It contains 6 electrons in$2 \mathrm{p}$orbital, 6 electrons in$3 \mathrm{p}$orbital and 5 electrons in$4 \mathrm{p}$orbital. Which of these electron experiences the lowest effective nuclear charge? Solution: Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases. Among p-orbitals,$4 \mathrm{p}$orbitals are farthest from the nucleus of bromine atom with$(+35)$charge. Hence, the electrons in the$4 \mathrm{p}$orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the$2 \mathrm{p}$and$3 \mathrm{p}$orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge. Question 63. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i)$\quad 2$s and$3 s$, (ii)$4 \mathrm{~d}$and$4 \mathrm{f}$, (iii)$3 \mathrm{~d}$and$3 \mathrm{p}$Solution: Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron$(\mathrm{s})$in it. (i) The electron(s) present in the$2 \mathrm{~s}$orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3 s orbital. (ii) 4 d will experience greater nuclear charge than$4 \mathrm{f}$since$4 \mathrm{~d}$is closer to the nucleus. (iii)$3 p$will experience greater nuclear charge since it is closer to the nucleus than$3 \mathrm{f}$. Question 64. The unpaired electrons in$\mathrm{Al}$and$\mathrm{Si}$are present in$3 \mathrm{p}$orbital. Which electrons will experience more effective nuclear charge from the nucleus? Solution: Nuclear charge is defined as the net positive charge experienced by an electron in a multielectron atom. The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of$(+14)$than aluminium, which has a nuclear charge of$(+13)$. Thus, the electrons in the$3 \mathrm{p}$orbital of silicon will experience a more effective nuclear charge than aluminium. Question 65. Indicate the number of unpaired electrons in: (a)$\mathrm{P}$(b)$\mathrm{Si}$(c)$\mathrm{Cr}$(d)$\mathrm{Fe}$(e)$\mathrm{Kr}$Solution: (a) Phosphorus (P): We know: Atomic number of$\mathrm{P}=15$The electronic configuration of$P$is: 1 s 22 s$22 \mathrm{p} 63 \mathrm{~s} 23 \mathrm{p} 3$The orbital picture of$\mathrm{P}$can be represented as: From the orbital picture, phosphorus has three unpaired electrons. (b) Silicon (Si): We know: Atomic number of$\mathrm{Si}=14$The electronic configuration of Si is: 1s 22 s 22263 s$23 p 2$From the orbital picture, silicon has two unpaired electrons. (c) Chromium (Cr): We know: Atomic number of$\mathrm{Cr}=24$The electronic configuration of$\mathrm{Cr}$is: 1 s$22 \mathrm{~s} 22 \mathrm{p} 63 \mathrm{~s} 23 \mathrm{p} 64 \mathrm{~s} 13 \mathrm{~d} 5$The orbital picture of chromium is: From the orbital picture, chromium has six unpaired electrons. (d) Iron (Fe): We know: Atomic number of$\mathrm{Fe}=26$The electronic configuration is: 1 s 22 s2 2 p 63 s 23 p 64 s 2336 The orbital picture of chromium is: From the orbital picture, iron has four unpaired electrons. (e) Krypton (Kr): We know: Atomic number of$\mathrm{Kr}=36$The electronic configuration is:$1 \mathrm{~s} 22 \mathrm{~s} 22 \mathrm{p} 63 \mathrm{~s} 23 \mathrm{p} 64 \mathrm{~s} 23 \mathrm{~d} 104 \mathrm{p} 6$The orbital picture of krypton is: Since all orbitals are fully occupied, there are no unpaired electrons in krypton. Question 66. (a) How many sub-shells are associated with$\mathrm{n}=4$? (b) How many electrons will be present in the sub-shells having ms value of$-\frac{1}{2}$for$\mathrm{n}=4$? Solution: (a) Given:$\mathrm{n}=4$For a given value of ‘$\mathrm{n}$‘, ‘ ‘l’ can have values from zero to$(\mathrm{n}-1)$. Therefore,$=0,1,2,3$Thus, four sub-shells are associated with$\mathrm{n}=4$, which are$\mathrm{s}, \mathrm{p}, \mathrm{d}$and$\mathrm{f}$. (b) Given:$\mathrm{n}=4$Number of orbitals in the$\mathrm{n}$th shell$=\mathrm{n}^{2}$For$\mathrm{n}=4$Number of orbitals$=16$If each orbital is taken fully, then it will have 1 electron with ms value of. Therefore, Number of electrons with$\mathrm{m}_{\mathrm{s}}$value of$(-12)$is 16 . Also Read, Class 11 Chemistry Notes Free PDF Download. Class 11 Chemistry Book Chapterwise Free PDF Download Class 11 Chemistry Exemplar Chapterwise Free PDF Download If you have any Confusion related to NCERT Solutions for Class 11 Chemistry chapter 2 Structure of Atom PDF then feel free to ask in the comments section down below. To watch Free Learning Videos on Class 11 Chemistry by Kota’s top IITian’s Faculties Install the eSaral App Structure of Atom Class 11 Questions and Answers-Important for Exams Get Structure of Atom important questions and answers for class 11 Chemistry exams. View the Important Question bank for Class 11 & 12 Chemistry developed by expert faculties from Kota. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapter.Learn all the concepts of Structure class 11 chemistry of Atom Chapter through these questions and answers.Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. You can access free study material for all three subject’s Physics, Chemistry and Mathematics.Click Here for Detailed Notes of any chapter. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. We have transformed classroom in such a way that a student can study anytime anywhere. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Visit eSaral Website to download or view free study material for JEE & NEET. Also get to know about the strategies to Crack Exam in limited time period. Q. How many protons and neutrons are present in the following nuclei? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) 6, 6 (ii) 26, 30 (iii) 38, 50 (iv) 92, 143. Q. The diameter of zinc atom is 2.6 A. Calculate(i) the radius of zinc atom in pm(ii) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Q. A certain particle carries$2.5 \times 10^{-16} \mathrm{C}$of static electric charge. Calculate the number of electrons present in it. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Charge carried by one electron$=1.6022 \times 10^{-19} \mathrm{C}$$\therefore \quad Electrons present in particle carrying2.5 \times 10^{-16} \mathrm{C} charge =\frac{2.5 \times 10^{-16}}{-1.6022 \times 10^{-19}}=1560 Q. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge ?(a) 2 s and 3 s, (b) 4 d and 4 f, (c) [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) 2 s is closer to the nucleus than 3 s. Hence 2 swill experience larger effective nuclear charge.(ii) 4 d (iii) 3 p (for same reasons). Q. The unpaired electrons in A l and S i are present in 3 p orbital. Which electron will experience more effective charge from the nucleus? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Silicon has greater nuclear charge (+14) than aluminium (+13) . Hence, the unpaired 3 p electron in case of silicon will experience more effective nuclear charge. Q. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven, (b) amber light. from traffic signal, (c) radiation from FMradio, (d) cosmic rays from outer space and (e) X -rays. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. The increasing order of frequency is :Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < X-rays < cosmic rays from outer space. Q. The energy associated with the first orbit of hydrogen atom is -2.17 \times 10^{-18} \mathrm{J} / atom. What is the energy associated with the fifth orbit? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. E_{n}=\frac{E_{1}}{n^{2}} ; E_{5}=\frac{-2.17 \times 10^{-18}}{(5)^{2}} \mathrm{J} / atom=-8.68 \times 10^{-20} \mathrm{J} / atom. Q. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 \times 10^{24}, calculate the power of this laser. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. E=N h v=N h \frac{c}{\lambda} \quad\left(\because v=\frac{c}{\lambda}\right)$$=\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{J} \mathrm{s}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{m}\right)}$$=3.3 \times 10^{6} J Q. What is meant by quantization of energy ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Quantization of energy means the energy of energy levels can have some specific values of energy and not all the values. Q. Write the mathematical expression by which the energies of various stationary states in the hydrogen atom can be calculated. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. \mathrm{E}_{\mathrm{n}}=-\frac{1312 \mathrm{kJ} \mathrm{mol}^{-1}}{\mathrm{n}^{2}} Q. How many subshells associated with n=4 ? How many electrons will be present in the subshell having m_{s} value of-1 / 2 for n=4 ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (a) Subshells in n=4 are$4 s(l=0), 4 p(l=1), 4 d(l=2), 4 f(l=3)$The number of electrons having m_{s}=-\frac{1}{2} in n=4 will be 16 Q. Explain the meaning of the symbol 4 d^{6} ? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. It means that 4d sub-shell has 6 electrons. 4 represents fourth energy shell and d is a sub-shell and 6 electrons are present in d orbitals of sub-shell. Q. How does Heisenberg’s Uncertainty principle support concept of orbital? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. According to Heisenberg’s Uncertainty principle ‘the exact position of electron cannot be determined, i.e., the exact path of electron cannot be determined but we can determine the region or space where electron spends most of its time, and this region or space is called orbital. Q. Explain why the uncertainty principle is significant only for the motion of subatomic particles but is negligible for the macroscopic objects. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. The energy of photon is sufficient to disturb a subatomic particle so that there is uncertainty in the measurement of position and momentum of the subatomic particle. However, the energy is insufficient to disturb a macroscopic object. Q. Can we apply Heisenberg’s uncertainty principle to a stationary electron ? Why or why not? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. No, because, velocity = 0 and position can be measured accurately. Q. Which of the four quantum numbers \left(n, l, m_{l}, m_{s}\right) determine:(i) the energy of an electron in a hydrogen atom and multi electron atoms(ii) the size of an orbital(iii) the shape of an orbital(iv) the orientation of an orbital in space ? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) n (ii) n (iii) l (iv) m_{l} Q. What physical meaning is attributed to the square of the absolute value of wave function \left|\psi^{2}\right| ? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. It measures the electron probability density at a point in an atom. Q. How many electrons in an atom may have the following quantum number ?(i) n=4, m_{s}=+1 / 2(ii) n=3, l=0 [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) n=4, m_{\mathrm{s}}=+1 / 2=16 electrons(ii) n=3, l=0=2 Q. For each of the following pair of hydrogen orbitals, indicate which is higher in energy :(i) 1 s, 2 s(ii) 2 p, 3 p(iii) 3 d_{x y}, 3 d_{y z}(iv) 3 s, 3 d(v) 4 f, 5 s [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) 2 s(ii) 3 p(iii) both have same energy(iv) both have same energy(v) 5 \mathrm{s} Q. The unpaired electrons in A l and S i are present in 3 p orbital. Which electrons will experience more effective nuclear charge from the nucleus? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Si Q. An electron is in one of the 3 d orbitals. Give the possible values of n, l, and m_{l} for this electron.[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. \mathrm{n}=3,1=2, \mathrm{m}_{1}=+2,+1,0,-1,-2 Q. List the quantum numbers ( l and m_{l} ) for electrons in 3d orbital. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 1=2, and \mathrm{m}_{1}=+2,+1,0,-1,-2 Q. Give the number of electrons in the species H_{2}^{+}, H_{2} andO_{2}^{+} Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Number of electrons :H_{2}^{+}=1 ; H_{2}=2 ; \quad O_{2}^{+}=15 Q. What atoms are indicated by the following configurations:(i) [\mathrm{He}] 2 \mathrm{s}^{1}(ii) [\mathrm{Ne}] 3 \mathrm{s}^{2} 3 \mathrm{p}^{3} \quad(iii) [\mathrm{Ar}] 4 \mathrm{s}^{2} 3 \mathrm{d}^{1} [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) L i(ii) P \quad(iii) S c Q. 2 \times 10^{8} atoms of carbon are arranged side by side. Calculate the radius of carbon atoms if length of this arrangement is 2.4 \mathrm{cm} Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. An atom will cover length equal to its diameterDiameter of carbon atom =\frac{2.4}{2 \times 10^{8}}$$=1.2 \times 10^{-8} \mathrm{cm}=1.2 \times 10^{-10} \mathrm{m}$Radius of carbon atom$=\frac{1.2 \times 10^{-10}}{2}=0.6 \times 10^{-10} \mathrm{m}$$=0.06 \times 10^{-9} \mathrm{m}=0.06 \mathrm{nm} Q. The wavelength range of the visible spectrum extends from = violet (400 \mathrm{nm}) to red (750 \mathrm{nm}) . Express their wave-lengths in frequencies (H z) .\left(1 n m=10^{-9} m\right) [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. For violet light v=7.50 \times 10^{14} \mathrm{Hz}For red light v=4.00 \times 10^{14} \mathrm{Hz} Q. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol for the ion. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. since the ion carries 3 units of positive charge, it will have 3 electrons less than the number of protons.Let number of electrons =xNo. of protons =x+3No. of neutrons =x+\frac{x \times 30.4}{100}$$=x+0.304 x=1.304 x$Now, No. of protons$+$No. of neutrons$=56$$x+3+1.304 x=56$$2.304 x=53$$x=\frac{53}{2.304}=23No. of electrons =23, No. of protons =23+3=26No. of neutrons =56-26=30 Q. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. We know, Mass number = No. of protons + No. of neutrons =81i.e., p+n=81 Let number of protons =xNumber of neutrons =x+\frac{x \times 31.7}{100}=1.317 x$$\therefore \quad x+1.317 x=81$$x=\frac{81}{2.317}=34.96=35Symbol =\begin{array}{l}{81} \\ {35}\end{array} B r Q. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Since the ion carries one unit of negative charge, it means that in the ion the number of electrons is one more than the number of protons.Total number of electrons and neutrons in the ion = 37 + 1 = 38Let the number of electrons in the ion be = xNumber of neutrons in the =\frac{x \times 111.1}{100}=1.111 x$$=x+1.111 x=2.111 x$$2.111 x=38$x=\frac{38}{2.111}=18$\therefore Number of electrons in the ion =18Number of protons in the ion =18-1=17Thus, atomic number of the elements is 17 which corresponds to chlorine.\therefore Symbol of the ion is _{17}^{37} \mathrm{Cl}^{-} Q. The threshold frequency v_{o} for a metal is 7.0 \times 10^{14} \mathrm{s}^{-1} Calculate the kinetic energy of an electron emitted when radiation of v=1.0 \times 10^{15} \mathrm{s}^{-1} hits the metal. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. K . E .=\frac{1}{2} m v^{2}=\frac{1}{2} m_{e} v^{2}=h\left(v-v_{0}\right)$$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(1.0 \times 10^{15} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)$$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(10 \times 10^{14} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)$$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{14} \mathrm{s}^{-1}\right)=1.988 \times 10^{-19} \mathrm{J}$Q. The Vividh Bharti station of All India Radio, Delhi broadcasts on a frequency of 1,368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong ? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. The wavelength,$\lambda,$is equal to$c / v$where$^{t} c^{\prime}$is the speed of electromagnetic radiation in vacuum and ‘$v^{\prime}$is the frequency. Substituting the given values, we have,$\lambda=\frac{3.00 \times 10^{8} \mathrm{ms}^{-1}}{1368 \mathrm{kHz}}$$=\frac{3.00 \times 10^{8} \mathrm{ms}^{-1}}{1368 \times 10^{3} \mathrm{s}^{-1}}=219.3 \mathrm{m}This is a characteristic radiowave wavelength. Q. Emission transitions in the Paschen series end at orbit n=3 and start from an orbit n and can be represented asv=3.29 \times 10^{15}(\mathrm{Hz})\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) Calculate the value of n if the transition is obtained at 1285 \mathrm{nm} . Find the region of spectrum. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. v=3.29 \times 10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) s^{-1}$$\lambda=1285 \mathrm{nm}$$\therefore \quad v=\frac{c}{\lambda}=\frac{3.0 \times 10^{8} \mathrm{ms}^{-1}}{1285 \times 10^{-9} \mathrm{m}}$$=2.33 \times 10^{14} s^{-1}$$2.33 \times 10^{14}=3.29 \times 10^{15}\left(\frac{1}{9}-\frac{1}{n^{2}}\right)or \quad \frac{1}{9}-\frac{1}{n^{2}}=\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}}=0.0708$$|-\frac{1}{n^{2}}=0.0708-\frac{1}{9}=-0.0403$$\frac{1}{n^{2}}=0.0403 or n^{2}=24.82$$\therefore \quad n=5$Q. Electrons are emitted with zero velocity from a metal surface. when it is exposed to radiation of wavelength$6,800$A. Calculate threshold frequency$\left(v_{0}\right)$and work function$\left(W_{0}\right)$of the metal. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol.$h v=h \theta_{\circ}+\frac{1}{2} m v^{2} \quad$since$v=0$$h v=h v_{o}+0$$h v=h v_{0} \Rightarrow v=v_{0}$$v=\frac{c}{\lambda} \Rightarrow v=\frac{3 \times 10^{8} \mathrm{ms}^{-1}}{6800 \times 10^{-10} \mathrm{m}}=\frac{300}{86} \times 10^{14}=4.41 \times 10^{14} \mathrm{s}^{-1}$$v=v_{0}=4.41 \times 10^{14} \mathrm{s}^{-1}$$W_{o}=h v_{o}=6.626 \times 10^{-34} \mathrm{Js} \times 4.41 \times 10^{14} \mathrm{s}^{-1}=2.91 \times 10^{-19} \mathrm{J} Q. Calculate the energy associated with the first orbit of H e^{+} What is the radius of this orbit? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Energy of electron in nth orbit isE_{n}=-\frac{13.595}{n^{2}} Z^{2} e V atom ^{-1}For first orbit of H e^{+}, Z=2, n=1$$E_{1}=-\frac{13.595 \times(2)^{2}}{1^{2}}=-54.380 e V$Radius of nth orbit is,$r_{n}=\frac{0.529 n^{2}}{Z} A$Q. Calculate the wave number for the longest wavelength transition in Balmer series of atomic hydrogen. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. The longest wavelength corresponds to minimum energy$(\Delta E)$transition. For Balmer series this transition is from$n_{2}=3$to$n_{1}=2$$v=109677\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \mathrm{cm}^{-1}$$=109677\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right] \mathrm{cm}^{-1}$$=15233 \mathrm{cm}^{-1}=1.5233 \times 10^{4} \mathrm{cm}^{-1}or \quad=1.5233 \times 10^{6} \mathrm{m}^{-1} Q. Show that the circumference of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron moving around the orbit. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. According to Bohr postulate of angular momentum,.$m v r=n \frac{h}{2 \pi} \quad \text { or } \quad 2 \pi r=n \frac{h}{m v} \quad \ldots .(\mathrm{i})$Substituting this value in equation (i), we get 2 \pi r=n \lambdaThus, the circumference (2 \pi r) of the Bohr orbit for hydrogen atoms is an integral multiple of de-Broglie wavelength. Q. Calculate de-Broglie wavelength of an electron (mass =9.1 \times 10^{-31} \mathrm{kg} ) moving at 1 \% speed of light. \left(h=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}\right) [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. We know that \lambda=\frac{h}{m v}$$m=9.1 \times 10^{-31} \mathrm{kg}, h=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$$v=1 \% of speed of light =\frac{1 \times 3.0 \times 10^{8}}{100} m s^{-1}$$=3.0 \times 10^{6} \mathrm{ms}^{-1}\left(\because \text { speed of light }=3.0 \times 10^{8} \mathrm{ms}^{-1}\right)$Q. The mass of an electron is$9.1 \times 10^{-31} \mathrm{kg} .$If its K.E. is$3.0 \times 10^{-25} \mathrm{J},$calculate its wavelength. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol.$K . E .=\frac{1}{2} m v^{2}$$\therefore v=\sqrt{\frac{2 K \cdot E}{m}}=\sqrt{\frac{2 \times 3.0 \times 10^{-25} \mathrm{J}}{9.1 \times 10^{-31} \mathrm{kg}}=812} \mathrm{ms}^{-1}$$\left(1, J=1 k g m^{2} s^{-2}\right)$By de-Broglie equation,$\lambda=\frac{h}{m v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{kg}\right)\left(812 \mathrm{ms}^{-1}\right)}$Q. (i) How many sub-shells are associated with$n=4$?(ii) How many electrons will be present in the sub-shells having$m_{s}$value of$-1 / 2$for$n=4 ?$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i)$n=4, l=0,1,2,3,(4 \text { subshells, viz, } s, p, d \text { and } f)$(ii) No. of orbitals in 4 th shell$=n^{2}=4^{2}=16$Each orbital has one electron with$m_{s}=-1 / 2 .$Hence, therewill be 16 electrons with$m_{s}=-1 / 2$Q. Calculate the uncertainty in the velocity of a wagon of mass 2000 kg whose position is known to an accuracy of \pm10$\mathrm{m}$[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Mass of wagon$=2000 \mathrm{kg}$Uncertainty in position,$\Delta x=\pm 10 \mathrm{m}$According to Heisenberg uncertainty principle$\Delta x \times \Delta p=\frac{h}{4 \pi}$or$\Delta x \times \Delta v=\frac{h}{4 \pi m}$or$\quad \Delta v=\frac{h}{4 \pi m \Delta x}$Q. If the position of the electron is measured within an accuracy of$\pm 0.002 n m,$calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is$h / 4 \pi \times 0.05 n m,$is there any problem in defining this value. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol.$\Delta x=0.002 n m=2 \times 10^{-3} n m=2 \times 10^{-12} m$$\Delta x \times \Delta p=\frac{h}{4 \pi}$$\therefore \quad \Delta p=\frac{h}{4 \pi \Delta x}=\frac{6.626 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{4 \times 3.14 \times\left(2 \times 10^{-12} \mathrm{m}\right)}$$=2.638 \times 10^{-23} \mathrm{kg} \mathrm{ms}^{-1}Actual momentum =\frac{h}{4 \pi \times 0.05 \mathrm{nm}}=\frac{h}{4 \pi \times 5 \times 10^{-11} \mathrm{m}}It cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty. Q. Using the s, p, d notations, describe the orbital with the following quantum numbers : [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Q. How many sub-shells are associated with n=4 ? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. For n=4, l can have values 0,1,2,3 . Thus, there are four sub-shells in n=4 energy level. These four sub-shells are 4 s, 4 p, 4 d and 4 f Q. Indicate the number of unpaired electron in(i) P(\text { ii }) Si ( iii) C r(\text { iv }) Fe, and (v) K r [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. i) \quad_{15} P=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1}No. of unpaired electron =3(ii) \quad_{14} S i=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1}No. of unpaired electron =2(iii) \quad_{24} C r=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}No. of unpaired electrons =6(iv) \quad 26 F e=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}No. of unpaired electrons =4(\text { in } 3 d)(v) \quad 36 K r= Noble gas. All orbitals are filled.Unpaired electron =0 Q. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. No. of protons = Number of electrons =29Electronic configuration=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{10} Q. Consider the following electronic configurations :(i) 1 s^{2} 2 s^{1}(ii) 1 s^{2} 3 s^{1}(a) Name the element corresponding to (i)(b) Does (ii) correspond to the same or different element?(c) How can (ii) be obtained from (i) ?(d) Is it easier to remove one electron from (ii) or (i) ? Explain. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (a) The element corresponding to ( i ) is Lithium (Li).(b) This electronic configuration represents the same element in the excited state.(c) By supplying energy to the element when the electron jumps from the lower energy 2 s -orbital to the higher energy 3 s orbital.(d) It is easier to remove an electron from (ii) than from (i) since in the former case the electron is present in a 3 s -orbital which is away from the nucleus and hence is less strongly attracted by the nucleus than an electron in the 2 s -orbital. Q. What atoms are indicated by the following configuration? Are they in the ground state or excited state ?(i) 1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{1}(ii) 1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}(iii) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1}(iv) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d^{1}(v) [A r] 3 d^{5} 4 s^{2} Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) 1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{1} . Atomic number is 9 and it is configuration of fluorine in ground state.(ii) 1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1} . Atomic number is 6 and it is configuration of carbon in excited state.(iii) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} . Atomic number is 14 and it is configuration of silicon in ground state.(iv) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d^{1} . Atomic number is 15 and it is configuration of phosphorus in excited state.(v) [A r] 3 d^{5} 4 s^{2} . Atomic number is 25 and it is configuration of manganese in ground state. Q. An atom has 2 electrons in the first (K) shell, 8 electrons in the second (L) shell and 2 electrons in the third (M) shell. Give its electronic configuration and find out the following:(i) Atomic number(ii) Total number of principal quantum numbers(iii) Total number of sub-levels(iv) Total number of s -orbitals(v) Total number of p -electrons. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. The electronic configuration of the atom is : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}(i) Atomic number =2+2+6+2=12(ii) Number of principal quantum numbers =3(iii) Number of sub-levels =4(1 s, 2 s, 2 p, 3 s)(iv) Number of s -orbitals =3(1 s, 2 s, 3 s)(v) Total number of p -electrons =6 Q. Account for the following:(i) The expected electronic configuration of copper is [A r] 3 d^{9} 4 s^{2} but actually it is [A r] 3 d^{10} 4 s^{1}(ii) In building up of atoms, the filling of 4 s orbitals occurs before 3 d- orbitals. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) It is because half filled and completely filled orbitals are more stable.(ii) 4 s orbital has lower energy than 3 d orbital, therefore, it is filled before 3 d orbital. Q. (i) Calculate the total number of electrons present in 1 mol of methane.(ii) Find (a) the total number and (b) total mass of neutrons in 7 m g of ^{14} C. (Assume mass of a neutron$\left.=1.675 \times 10^{-27} \mathrm{kg}\right)$(iii) Find (a) total number and (b) total mass of protons in 34 m g of N H_{3} at STP.Will the answer change if the temperature and pressure are changed? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) 1 molecule of methane contains =6+4=10 electrons, 6.022 \times 10^{23} molecules of methane (1 \mathrm{mol}) contain electrons =$$10 \times 6.022 \times 10^{23}=6.022 \times 10^{24}$(ii) One atom of$^{14} C$contains$=8$neutrons,$6.022 \times 10^{23}$atomsof carbon weight$=14 \mathrm{g}$$14 g of C contain =6.022 \times 10^{23} atoms=6.022 \times 10^{23} \times 8 neutrons$7.0 \times 10^{-3} g \text { of } C \text { contains }$=\frac{6.022 \times 10^{23} \times 8 \times 7.0 \times 10^{-3}}{14}=2.409 \times 10^{21}Mass of 1 neutron =1.675 \times 10^{-27} \mathrm{kg}Mass of 2.409 \times 10^{21} neutrons=1.675 \times 10^{-27} \times 2.409 \times 10^{21}$$=4.035 \times 10^{-6} \mathrm{kg}$(iii) No. of protons in 1 molecule of$N H_{3}$$=7+1+1+1=10 \text { protons }$$17 g$of$N H_{3}$contains$=6.022 \times 10^{23}$molecules$=6.022 \times 10^{23} \times 10$protons$=6.022 \times 10^{24}$protons$34 \times 10^{-3} g$of$N H_{3}$contains$=\frac{6.022 \times 10^{24} \times 34 \times 10^{-3}}{17}=1.204 \times 10^{22}$Mass of 1 proton$=1.67 \times 10^{-27} \mathrm{kg}$Mass of$1.204 \times 10^{22}$protons$=1.67 \times 10^{-27} \times 1.204 \times 10^{22}$$=2.01 \times 10^{-5} \mathrm{kg}No answer with not change by changing temp. and pres., Q. When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1} . What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. – Energy of a photon of radiation of wavelength 300 nm will beE=h v=h \frac{c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(300 \times 10^{-9} \mathrm{m}\right)}$=6.626 \times 10^{-19} \mathrm{J}$\therefore Energy of 1 mole of photons=\left(6.626 \times 10^{-19} \mathrm{J}\right) \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)$$=3.99 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$As$\quad E=E_{0}+K . E .$of photoelectrons emitted.$\therefore$Miniumu energy$\left(E_{0}\right)$required to remove 1 mole of electrons from sodium$=E-K . E$$=(3.99-1.68) 10^{5} J m o l^{-1}$$=2.31 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$$\therefore Minimum energy required to remove one electron. Q. Following results were observed when sodium metal is irradiated with different wavelengths. Calculate (a) thres-hold wavelength and (b) Planck’s constant. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Suppose threshold wavelength =\lambda_{0} n m=\lambda_{0} \times 10^{-9} \mathrm{m}Then h\left(v-v_{0}\right)=\frac{1}{2} m v^{2} \quad or \quad h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m v^{2}Substituting the given results of the three experiments, we get\frac{h c}{10^{-9}}\left(\frac{1}{500}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(2.55 \times 10^{6}\right)^{2}$$\frac{h c}{10^{-9}}\left(\frac{1}{450}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(4.35 \times 10^{6}\right)^{2}$$\frac{h c}{10^{-9}}\left(\frac{1}{400}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(5.20 \times 10^{6}\right)^{2}Dividing equation (ii) by equation (i), we get\frac{\lambda_{0}-450}{450 \lambda_{0}} \times \frac{500 \lambda_{0}}{\lambda-500}=\left(\frac{4.35}{2.55}\right)^{2}or \quad \frac{\lambda_{0}-450}{\lambda_{0}-500}=\frac{450}{500}\left(\frac{4.35}{2.55}\right)^{2}=2.619or \lambda_{0}-450=2.619 \lambda-1309.5or \quad 1.619 \lambda_{0}=859.5 \quad \therefore \quad \lambda_{0}=531 \mathrm{nm}Substituting this value in equation (iii), we get\frac{h \times\left(3 \times 10^{8}\right)}{10^{-9}}\left(\frac{1}{400}-\frac{1}{531}\right)=\frac{1}{2}\left(9.11 \times 10^{-31}\right)\left(5.20 \times 10^{6}\right)^{2}or \quad h=6.66 \times 10^{-34} \mathrm{Js} Q. For which hydrogen like ion the wavelength difference between the first lines of the Lyman and Balmer series is equal to 59.3 nm Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. For a spectral line,\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]For Lyman line,\frac{1}{\lambda_{\mathrm{Lyman}}}=R Z^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3 R Z^{2}}{4}or \quad \lambda_{L y \operatorname{man}}=\frac{4}{3 R Z^{2}}For Balmer line,\frac{1}{\lambda_{\text {Balmer }}}=R z^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5 R Z^{2}}{36}or \quad \lambda_{\text {Balmer }}=\frac{36}{5 R Z^{2}}$$\lambda_{\text {Balmer }}-\lambda_{\text {Lyman }}=\frac{36}{5 R Z^{2}}-\frac{4}{3 R Z^{2}}=59.3 \times 10^{-7}$$=\frac{1}{R Z^{2}}\left[\frac{36}{5}-\frac{4}{3}\right]=59.3 \times 10^{-7}$$=\frac{1}{109677.8 Z^{2}}\left[\frac{108-20}{15}\right]=59.3 \times 10^{-7}$or$\quad Z^{2}=\frac{1}{109677.8} \times \frac{88}{15 \times 59.3 \times 10^{-7}}=9$or$\quad Z=3 .$This corresponds to$L i^{2+}$ion. Q. Calculate the velocity$\left(\mathrm{cms}^{-1}\right)$of an electron placed in third orbit of the hydrogen atom. Also calculate the number of revolutions per second that this electron makes around the nucleus. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Velocity of electron is given as :$v=\frac{2 \pi e^{2}}{n h}$Now,$e=4.8 \times 10^{-10}$e.s.u.,$n=3, h=6.63 \times 10^{-27}$erg sec$\therefore \quad v=\frac{2 \times 22 \times\left(4.8 \times 10^{-10}\right)^{2}}{7 \times 3 \times 6.63 \times 10^{-27}}=7.27 \times 10^{7} \mathrm{cm} \mathrm{s}^{-1}$No. of revolutions per second$=\frac{\text { Velocity }}{\text { Circumference }}=\frac{v}{2 \pi r}$but$r=\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}}$$\therefore No. of revolutions =\frac{2 \pi e^{2}}{n h} \times \frac{4 \pi^{2} m e^{2}}{2 \pi n^{2} h^{2}}=\frac{4 \pi^{2} m e^{4}}{n^{3} h^{3}}$$=\frac{4 \times\left(\frac{22}{7}\right)^{2} \times\left(9.1 \times 10^{-28}\right) \times\left(4.8 \times 10^{-10}\right)^{4}}{3^{3} \times\left(6.62 \times 10^{-27}\right)^{3}}$$=2.42 \times 10^{14} Q. Calculate the wavelength in Angstroms of the photon that is emitted when an electron in Bohr orbit n=2 returns to the orbit n=1 in the hydrogen atom. The ionisation potential of the ground state of hydrogen atom is 2.17 \times 10^{-11} ergs per atom. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Q. The angular momentum of an electron in Bohr’s orbit of hydrogen atom is 4.22 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1} . Calculate the wavelength of the spectral line when the electron falls from this level to the next lower level. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Angular momentum (m v r)=n \frac{h}{2 \pi}$$=4.22 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$(Given)$\therefore \quad n=4.22 \times 10^{-34} \times \frac{2 \pi}{h}$$=\frac{2 \times 4.22 \times 10^{-34} \times 3.14}{6.626 \times 10^{-34}}=4When the electron jumps from n=4 to n=3, the wavelength of the spectral line can be calculated as follows:\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$$=109,677 \mathrm{cm}^{-1}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$$=109,677 \times\left(\frac{1}{9}-\frac{1}{16}\right)=109,677 \times \frac{7}{144} \mathrm{cm}^{-1}or \lambda=\frac{144}{109,677 \times 7} \mathrm{cm}=1.88 \times 10^{-4} \mathrm{cm} Q. (i) Write the electronic configuration of C u^{+} ion (Z=29).(ii) Calculate the de-Broglie wavelength of a milligram sized object moving with 1 \% speed of light.Planck’s constant (h)=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}Velocity of light (c)=3.0 \times 10^{8} \mathrm{ms}^{-1} Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) \quad C u^{+}(29) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{10}(ii) \quad m=1 m g \quad v=1 \% \times 3 \times 10^{8} m s^{-1}$$m=10^{-6} \mathrm{kg} \quad v=\frac{1}{100} \times 3 \times 10^{8} \mathrm{ms}^{-1}=3 \times 10^{6} \mathrm{ms}^{-1}$$\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{10^{-6} \mathrm{kg} \times 3 \times 10^{6} \mathrm{ms}^{-1}}=2.21 \times 10^{-34} \mathrm{m} Q. Give the electronic configuration of the following ions: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. During the formation of cations, electrons are lost while in the formation of anions, electrons are added to the valence shell. The number of electrons added or lost is equal to the numerical value of the charge present on the ion. Following this general concept, we can write the electronic configurations of all the ions given in the question.(i) \quad C u^{2+}=_{29} C u-2 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{1}-2 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{9}(ii) \quad C r^{3+}=_{24} C r-3 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}-3 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}(iii) \quad F e^{2+}=_{26} F e-2 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}-2 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$$F e^{3+}=26 F e-3 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}$(iv)$\quad H^{-}=_{1} H+1 e^{-}=1 s^{1}+1 e^{1}=1 s^{2}$(v)$\quad S^{2-}=_{16} S+2 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{1} 3 p_{z}^{1}+2 e^{-}$$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{2} 3 p_{z}^{2}\$