Sol. (i) 6, 6 (ii) 26, 30 (iii) 38, 50 (iv) 92, 143.

Sol.

Sol. Charge carried by one electron $=1.6022 \times 10^{-19} \mathrm{C}$

$\therefore \quad$ Electrons present in particle carrying

$2.5 \times 10^{-16} \mathrm{C}$ charge $=\frac{2.5 \times 10^{-16}}{-1.6022 \times 10^{-19}}=1560$

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).

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.

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.

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.

Sol. $E=N h v=N h \frac{c}{\lambda} \quad\left(\because v=\frac{c}{\lambda}\right)$

$=\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{J} \mathrm{s}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{m}\right)}$

$=3.3 \times 10^{6} J$

Sol. Quantization of energy means the energy of energy levels can have some specific values of energy and not all the values.

Sol. $\mathrm{E}_{\mathrm{n}}=-\frac{1312 \mathrm{kJ} \mathrm{mol}^{-1}}{\mathrm{n}^{2}}$

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

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.

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.

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.

Sol. No, because, velocity = 0 and position can be measured accurately.

Sol. (i) n (ii) n (iii) l (iv) $m_{l}$

Sol. It measures the electron probability density at a point in an atom.

Sol. (i) $n=4, m_{\mathrm{s}}=+1 / 2=16$ electrons

(ii) $n=3, l=0=2$

Sol.

(i) $2 s$

(ii) $3 p$

(iii) both have same energy

(iv) both have same energy

(v) $5 \mathrm{s}$

Sol. Si

Sol. $\mathrm{n}=3,1=2, \mathrm{m}_{1}=+2,+1,0,-1,-2$

Sol. $1=2,$ and $\mathrm{m}_{1}=+2,+1,0,-1,-2$

Sol. Number of electrons :

$H_{2}^{+}=1 ; H_{2}=2 ; \quad O_{2}^{+}=15$

Sol. (i) $L i$

(ii) $P \quad$

(iii) $S c$

Sol. An atom will cover length equal to its diameter

Diameter 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 $=\frac{1.2 \times 10^{-10}}{2}=0.6 \times 10^{-10} \mathrm{m}$

$=0.06 \times 10^{-9} \mathrm{m}=0.06 \mathrm{nm}$

Sol. For violet light $v=7.50 \times 10^{14} \mathrm{Hz}$

For red light $v=4.00 \times 10^{14} \mathrm{Hz}$

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 $=x$

No. of protons $=x+3$

No. of neutrons $=x+\frac{x \times 30.4}{100}$

$=x+0.304 x=1.304 x$

Now, No. of protons $+$ No. of neutrons $=56$

$x+3+1.304 x=56$

$2.304 x=53$

$x=\frac{53}{2.304}=23$

No. of electrons $=23,$ No. of protons $=23+3=26$

No. of neutrons $=56-26=30$

Sol. We know, Mass number $=$ No. of protons $+$ No. of neutrons $=81$

i.e., $p+n=81$ Let number of protons $=x$

Number of neutrons $=x+\frac{x \times 31.7}{100}=1.317 x$

$\therefore \quad x+1.317 x=81$

$x=\frac{81}{2.317}=34.96=35$

Symbol $=\begin{array}{l}{81} \\ {35}\end{array} B r$

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 = 38

Let the number of electrons in the ion be = x

Number 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 $=18$

Number of protons in the ion $=18-1=17$

Thus, atomic number of the elements is 17 which corresponds to chlorine.

$\therefore$ Symbol of the ion is $_{17}^{37} \mathrm{Cl}^{-}$

Sol. $K . E .=\frac{1}{2} m v^{2}=\frac{1}{2} m_{e} v^{2}=h\left(v-v_{0}\right)$

$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(1.0 \times 10^{15} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)$

$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(10 \times 10^{14} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)$

$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{14} \mathrm{s}^{-1}\right)=1.988 \times 10^{-19} \mathrm{J}$

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.

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 $n^{2}=24.82$

$\therefore \quad n=5$

Sol. $h v=h \theta_{\circ}+\frac{1}{2} m v^{2} \quad$ since $v=0$

$h v=h v_{o}+0$

$h v=h v_{0} \Rightarrow v=v_{0}$

$v=\frac{c}{\lambda} \Rightarrow v=\frac{3 \times 10^{8} \mathrm{ms}^{-1}}{6800 \times 10^{-10} \mathrm{m}}=\frac{300}{86} \times 10^{14}=4.41 \times 10^{14} \mathrm{s}^{-1}$

$v=v_{0}=4.41 \times 10^{14} \mathrm{s}^{-1}$

$W_{o}=h v_{o}=6.626 \times 10^{-34} \mathrm{Js} \times 4.41 \times 10^{14} \mathrm{s}^{-1}=2.91 \times 10^{-19} \mathrm{J}$

Sol. Energy of electron in nth orbit is

$E_{n}=-\frac{13.595}{n^{2}} Z^{2} e V$ 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,

$r_{n}=\frac{0.529 n^{2}}{Z} A$

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 $\quad=1.5233 \times 10^{6} \mathrm{m}^{-1}$

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 $2 \pi r=n \lambda$

Thus, the circumference $(2 \pi r)$ of the Bohr orbit for hydrogen atoms is an integral multiple of de-Broglie wavelength.

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 $=\frac{1 \times 3.0 \times 10^{8}}{100} m s^{-1}$

$=3.0 \times 10^{6} \mathrm{ms}^{-1}\left(\because \text { speed of light }=3.0 \times 10^{8} \mathrm{ms}^{-1}\right)$

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,

$\lambda=\frac{h}{m v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{kg}\right)\left(812 \mathrm{ms}^{-1}\right)}$

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, there

will be 16 electrons with $m_{s}=-1 / 2$

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 $\quad \Delta v=\frac{h}{4 \pi m \Delta x}$

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.

Sol.

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 $4 f$

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 $=0$

Sol. No. of protons $=$ Number of electrons $=29$

Electronic configuration

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

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.

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.

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 $=6$

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.

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, $6.022 \times 10^{23}$ atoms

of 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.,

Sol. – Energy of a photon of radiation of wavelength 300 nm will be

$E=h v=h \frac{c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(300 \times 10^{-9} \mathrm{m}\right)}$

\[=6.626 \times 10^{-19} \mathrm{J}\]

$\therefore$ 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.

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 $\quad \frac{\lambda_{0}-450}{\lambda_{0}-500}=\frac{450}{500}\left(\frac{4.35}{2.55}\right)^{2}=2.619$

or $\lambda_{0}-450=2.619 \lambda-1309.5$

or $\quad 1.619 \lambda_{0}=859.5 \quad \therefore \quad \lambda_{0}=531 \mathrm{nm}$

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 $\quad h=6.66 \times 10^{-34} \mathrm{Js}$

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.

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 $=\frac{2 \pi e^{2}}{n h} \times \frac{4 \pi^{2} m e^{2}}{2 \pi n^{2} h^{2}}=\frac{4 \pi^{2} m e^{4}}{n^{3} h^{3}}$

$=\frac{4 \times\left(\frac{22}{7}\right)^{2} \times\left(9.1 \times 10^{-28}\right) \times\left(4.8 \times 10^{-10}\right)^{4}}{3^{3} \times\left(6.62 \times 10^{-27}\right)^{3}}$

$=2.42 \times 10^{14}$

Sol.

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)

$\therefore \quad n=4.22 \times 10^{-34} \times \frac{2 \pi}{h}$

$=\frac{2 \times 4.22 \times 10^{-34} \times 3.14}{6.626 \times 10^{-34}}=4$

When 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 $\lambda=\frac{144}{109,677 \times 7} \mathrm{cm}=1.88 \times 10^{-4} \mathrm{cm}$

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) $\quad m=1 m g \quad v=1 \% \times 3 \times 10^{8} m s^{-1}$

$m=10^{-6} \mathrm{kg} \quad v=\frac{1}{100} \times 3 \times 10^{8} \mathrm{ms}^{-1}=3 \times 10^{6} \mathrm{ms}^{-1}$

$\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{10^{-6} \mathrm{kg} \times 3 \times 10^{6} \mathrm{ms}^{-1}}=2.21 \times 10^{-34} \mathrm{m}$

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) $\quad S^{2-}=_{16} S+2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{1} 3 p_{z}^{1}+2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{2} 3 p_{z}^{2}$