The threshold frequency for a certain metal
Question: The threshold frequency for a certain metal is $3.3 \times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, predict the cutoff voltage for the photoelectric emission. Solution: Threshold frequency of the metal, $v_{0}=3.3 \times 10^{14} \mathrm{~Hz}$ Frequency of light incident on the metal, $v=8.2 \times 10^{14} \mathrm{~Hz}$ Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$ Planck's constant, $h=6.626 \times 10^{-34} \ma...
Read More →Show that the products of the corresponding terms of the sequences
Question: Show that the products of the corresponding terms of the sequences$a, a r, a r^{2}, \ldots a r^{n-1}$ and $A, A R, A R^{2}, \ldots A R^{n-1}$ form a G.P, and find thecommon ratio. Solution: It has to be proved that the sequence, $a A, \operatorname{arAR}, a r^{2} A R^{2}, \ldots a r^{n-1} A R^{n-1}$, forms a G.P. $\frac{\text { Second term }}{\text { First term }}=\frac{\text { ar } A R}{a A}=r R$ $\frac{\text { Third term }}{\text { Second term }}=\frac{a r^{2} A R^{2}}{a r A R}=r R$ ...
Read More →ind the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, .
Question: ind the sum of the products of the corresponding terms of the sequences$2,4,8,16,32$ and $128,32,8,2, \frac{1}{2}$ Solution: Required sum $=2 \times 128+4 \times 32+8 \times 8+16 \times 2+32 \times \frac{1}{2}$ $=64\left[4+2+1+\frac{1}{2}+\frac{1}{2^{2}}\right]$ Here, $4,2,1, \frac{1}{2}, \frac{1}{2^{2}}$ is a G.P. First term, $a=4$ Common ratio, $r=\frac{1}{2}$ It is known that, $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$ $\therefore S_{5}=\frac{4\left[1-\left(\frac{1}{2}\right)^{5}\rig...
Read More →A 100 W sodium lamp radiates energy uniformly in all directions.
Question: A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere? Solution: Power of the sodium lamp,P= 100 W Wavelength of the emitted sodium light,= 589 nm = 589109m Plancks constant,h= 6.6261034Js Speed of ...
Read More →A 100 W sodium lamp radiates energy uniformly in all directions.
Question: A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere? Solution: Power of the sodium lamp,P= 100 W Wavelength of the emitted sodium light,= 589 nm = 589109m Plancks constant,h= 6.6261034Js Speed of ...
Read More →In an experiment on photoelectric effect,
Question: In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \mathrm{Vs}$. Calculate the value of Planck's constant. Solution: The slope of the cut-off voltage (V) versus frequency () of an incident light is given as: $\frac{V}{V}=4.12 \times 10^{-15} \mathrm{Vs}$ $V$ is related to frequency by the equation: $h v=e V$ Where, e= Charge on an electron = 1.61019C h= Plancks constant $\therefore h=e \time...
Read More →For the matrix, find the numbers a and b such that A2 + aA + bI = O
Question: For the matrix $A=\left[\begin{array}{ll}3 2 \\ 1 1\end{array}\right]$, find the numbers $a$ and $b$ such that $A^{2}+a A+b l=0$. Solution: $A=\left[\begin{array}{ll}3 2 \\ 1 1\end{array}\right]$ $\therefore A^{2}=\left[\begin{array}{ll}3 2 \\ 1 1\end{array}\right]\left[\begin{array}{ll}3 2 \\ 1 1\end{array}\right]=\left[\begin{array}{ll}9+2 6+2 \\ 3+1 2+1\end{array}\right]=\left[\begin{array}{ll}11 8 \\ 4 3\end{array}\right]$ Now, $A^{2}+a A+b I=0$ $\Rightarrow(A A) A^{-1}+a A A^{-1}+...
Read More →The energy flux of sunlight reaching the surface of the earth is
Question: The energy flux of sunlight reaching the surface of the earth is 1.388 103W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm. Solution: Energy flux of sunlight reaching the surface of earth,Φ= 1.388103W/m2 Hence, power of sunlight per square metre,P= 1.388103W Speed of light,c= 3108m/s Plancks constant,h= 6.6261034Js Average wavelength of photons present in sunlight, $\lambd...
Read More →What is the difference between distillation,
Question: What is the difference between distillation, distillation under reduced pressure and steam distillation ? Solution: The differences among distillation, distillation under reduced pressure, and steam distillation are given in the following table....
Read More →Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Question: Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$ Solution: The given sequence is $8,88,888,8888 \ldots$ This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as $S_{n}=8+88+888+8888+\ldots \ldots \ldots \ldots \ldots \ldots$ to $n$ terms $=\frac{8}{9}[9+99+999+9999+\ldots \ldots \ldots .$ to $n$ terms $]$ $=\frac{8}{9}\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\left(10^{4}-1\right)+\ldots \ldots . .\right.$ to $n$ terms $]$ $...
Read More →Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser.
Question: Monochromatic light of wavelength 632.8 nm is produced by ahelium-neon laser. The power emitted is 9.42 mW. (a)Find the energy and momentum of each photon in the light beam, (b)How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c)How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? Solution: Wavelength of the mo...
Read More →Describe the method, which can be used to separate two compounds
Question: Describe the method, which can be used to separate two compounds with different solubilities in a solvent S. Solution: Fractional crystallisation is the method used for separating two compounds with different solubilities in a solvent S. The process of fractional crystallisation is carried out in four steps. (a) Preparation of the solution:The powdered mixture is taken in a flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is j...
Read More →If, show that. Hence find.
Question: If $A=\left[\begin{array}{rr}3 1 \\ -1 2\end{array}\right]$, show that $A^{2}-5 A+7 I=O$. Hence find $A^{-1}$. Solution: $A=\left[\begin{array}{rr}3 1 \\ -1 2\end{array}\right]$ $A^{2}=A \cdot A=\left[\begin{array}{rr}3 1 \\ -1 2\end{array}\right]\left[\begin{array}{rr}3 1 \\ -1 2\end{array}\right]=\left[\begin{array}{ll}9-1 3+2 \\ -3-2 -1+4\end{array}\right]=\left[\begin{array}{rr}8 5 \\ -5 3\end{array}\right]$ $\therefore A^{2}-5 A+7 I$ $=\left[\begin{array}{rr}8 5 \\ -5 3\end{array}...
Read More →If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Question: If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ terms of a G.P. are $x, y$ and $z$, respectively. Prove that $x, y, z$ are in G.P. Solution: Letabe the first term andrbe the common ratio of the G.P. According to the given condition, $a_{4}=a r^{3}=x \ldots(1)$ $a_{10}=a r^{9}=y \ldots(2)$ $a_{16}=a r^{15}=z \ldots$ (3) Dividing $(2)$ by $(1)$, we obtain $\frac{y}{x}=\frac{a r^{9}}{a r^{3}} \Rightarrow \frac{y}{x}=r^{6}$ Dividing (3) by (2), we obtain $\frac{z}{y}=\fra...
Read More →Give a brief description of the principles of the following techniques taking an example in each case.
Question: Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography Solution: (a) Crystallisation Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds. Principle:It is based on the difference in the solubilites of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly sol...
Read More →The photoelectric cut-off voltage in a certain experiment is 1.5 V.
Question: The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? Solution: Photoelectric cut-off voltage,V0= 1.5 V Themaximum kinetic energy of the emitted photoelectrons is given as: $K_{e}=e V_{0}$ Where, e= Charge on an electron = 1.61019C $\therefore K_{e}=1.6 \times 10^{-19} \times 1.5$ $=2.4 \times 10^{-19} \mathrm{~J}$ Therefore, the maximum kinetic energy ofthe photoelectrons emitted in the given experiment is 2.4...
Read More →find the
Question: Let $A=\left[\begin{array}{ll}3 7 \\ 2 5\end{array}\right]$ and $B=\left[\begin{array}{ll}6 8 \\ 7 9\end{array}\right]$. Verify that $(A B)^{-1}=B^{-1} A^{-1}$ Solution: Let $A=\left[\begin{array}{ll}3 7 \\ 2 5\end{array}\right]$ We have, $|A|=15-14=1$ Now, $A_{11}=5, A_{12}=-2, A_{21}=-7, A_{22}=3$ $\therefore \operatorname{adj} A=\left[\begin{array}{rr}5 -7 \\ -2 3\end{array}\right]$ $\therefore A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A=\left[\begin{array}{rr}5 -7 \\ -2 3\end{a...
Read More →The work function of caesium metal is 2.14 eV.
Question: The work function of caesium metal is 2.14 eV. When light of frequency 6 1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a)maximum kinetic energy of the emitted electrons, (b)Stopping potential, and (c)maximum speed of the emitted photoelectrons? Solution: Work function of caesium metal, $\phi_{0}=2.14 \mathrm{eV}$ Frequency of light, $v=6.0 \times 10^{14} \mathrm{~Hz}$ (a)The maximum kinetic energy is given by the photoelectric effect as: $K=h ...
Read More →Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.
Question: Find a G.P. for which sum of the first two terms is $-4$ and the fifth term is 4 times the third term. Solution: Letabe the first term andrbe the common ratio of the G.P. According to the given conditions, $S_{2}=-4=\frac{a\left(1-r^{2}\right)}{1-r}$$\ldots(1)$ $a_{5}=4 \times a_{3}$ $a r^{4}=4 a r^{2}$ $\Rightarrow r^{2}=4$ $\therefore r=\pm 2$ From (1), we obtain $-4=\frac{a\left[1-(2)^{2}\right]}{1-2}$ for $r=2$ $\Rightarrow-4=\frac{a(1-4)}{-1}$ $\Rightarrow-4=a(3)$ $\Rightarrow a=\...
Read More →The work function of caesium metal is 2.14 eV.
Question: The work function of caesium metal is 2.14 eV. When light of frequency 6 1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a)maximum kinetic energy of the emitted electrons, (b)Stopping potential, and (c)maximum speed of the emitted photoelectrons? Solution: Work function of caesium metal, $\phi_{0}=2.14 \mathrm{eV}$ Frequency of light, $v=6.0 \times 10^{14} \mathrm{~Hz}$ (a)The maximum kinetic energy is given by the photoelectric effect as: $K=h ...
Read More →Give a brief description of the principles of the following techniques taking an example in each case.
Question: Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography Solution: (a) Crystallisation Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds. Principle:It is based on the difference in the solubilites of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly sol...
Read More →Find the
Question: Find the (a)maximum frequency, and (b)minimum wavelength of X-rays produced by 30 kV electrons. Solution: Potential of the electrons, $V=30 \mathrm{kV}=3 \times 10^{4} \mathrm{~V}$ Hence, energy of the electrons, $E=3 \times 10^{4} \mathrm{eV}$ Where, e= Charge on an electron = 1.61019C (a)Maximum frequency produced by the X-rays = The energy of the electrons is given by the relation: E=h Where, h= Plancks constant = 6.6261034Js $\therefore v=\frac{E}{h}$ $=\frac{1.6 \times 10^{-19} \t...
Read More →Find the inverse of each of the matrices (if it exists).
Question: Find the inverse of each of the matrices (if it exists). $\left[\begin{array}{ccc}1 0 0 \\ 0 \cos \alpha \sin \alpha \\ 0 \sin \alpha -\cos \alpha\end{array}\right]$ Solution: Let $A=\left[\begin{array}{ccc}1 0 0 \\ 0 \cos \alpha \sin \alpha \\ 0 \sin \alpha -\cos \alpha\end{array}\right]$. We have, $|A|=1\left(-\cos ^{2} \alpha-\sin ^{2} \alpha\right)=-\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=-1$ Now, $A_{11}=-\cos ^{2} \alpha-\sin ^{2} \alpha=-1, A_{12}=0, A_{13}=0$ $A_{21}=0, ...
Read More →Explain the terms Inductive and Electromeric effects.
Question: Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) $\mathrm{Cl}_{3} \mathrm{CCOOH}\mathrm{Cl}_{2} \mathrm{CHCOOH}\mathrm{ClCH}_{2} \mathrm{COOH}$ (b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} \cdot \mathrm{COOH}$ Solution: Inductive effect The permanent displacement of sigma ()...
Read More →Find the inverse of each of the matrices (if it exists).
Question: Find the inverse of each of the matrices (if it exists). $\left[\begin{array}{ccc}1 -1 2 \\ 0 2 -3 \\ 3 -2 4\end{array}\right]$ Solution: $\left[\begin{array}{ccc}1 -1 2 \\ 0 2 -3 \\ 3 -2 4\end{array}\right]$ Let $A=\left[\begin{array}{ccc}1 -1 2 \\ 0 2 -3 \\ 3 -2 4\end{array}\right]$. By expanding along $C_{1}$, we have: $|A|=1(8-6)-0+3(3-4)=2-3=-1$ Now, $A_{11}=8-6=2, A_{12}=-(0+9)=-9, A_{13}=0-6=-6$ $A_{21}=-(-4+4)=0, A_{22}=4-6=-2, A_{23}=-(-2+3)=-1$ $A_{31}=3-4=-1, A_{32}=-(-3-0)=...
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