For a certain metal, when monochromatic light of wavelength \(\lambda\) is incident, the stopping potential for photoelectrons is \(3V_0.\) When the same metal is illuminated by light of wavelength \(2\lambda,\) then the stopping potential becomes \(V_0.\) The threshold wavelength for photoelectric emission for the given metal is \(\alpha\lambda\). The value of \(\alpha\) is:
1. \(1\)
2. \(4\)
3. \(2\)
4. \(3\)
Subtopic:  Einstein's Photoelectric Equation |
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Light source having wavelength \(331 ~\text {nm}\) to generate photo-electrons whose stopping potential is \(0.2~\text{V}\). The work function of the used metal in the experiment is \(\alpha \times 10^{-19} ~\text{J} .\) The value of \(\alpha\) is:
\(\left({h}=6.62 \times 10^{-34} \text{J s}, {e}=1.6 \times 10^{-19}~\text{C} \text { and } {c}=3 \times 10^8~ \text{m/s}\right)\)
1. \(3.68\)
2. \(4.68\)
3. \(5.68\)
4. \(2.68\)
Subtopic:  Einstein's Photoelectric Equation |
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\(K_1\) and \(K_2\) be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength \(\lambda_1\) and \(\lambda_2\), respectively. If \(\lambda_1=2\lambda_2\) then the work function of material is given by:
1. \(K_2+2K_1\)
2. \(2K_2-K_1\)
3. \(K_1-2K_2\)
4. \(K_2-2K_1\)
Subtopic:  Einstein's Photoelectric Equation |
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The graph shows variations of stopping potential \(V_0\) with the frequency \(\nu\) of the incident radiation for three photosensitive metals \(X_1,X_2~ \text{and} ~X_3.\) Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?
                       
1. \(X_1\)
2. \(X_2\)
3. \(X_3\)
4. All the metals will give out photo electrons with same kinetic energies. 
Subtopic:  Einstein's Photoelectric Equation |
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A light wave described by \(E = 60[\sin(3\times 10^{15})t+\sin(12\times 10^{15})t]\) (in SI units) falls on a metal surface of work function \(2.8~\text{eV}\). The maximum kinetic energy of ejected photoelectron is (approximately): (in eV)
(\(h = 6.6\times 10^{-34}~\text{J.s}~\text{and}~e = 1.6\times 10^{-19}~\text{C}\).)
1. \(5.1\)
2. \(3.8\)
3. \(6.0\)
4. \(7.8\)
Subtopic:  Einstein's Photoelectric Equation |
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Light is incident on a metallic plate having work function \(110\times 10^{20} ~\text {J}.\) If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is: (in rad/s)
 \((h=6.63\times 10^{-34} ~\text{J.s})\)
1. \(1.04\times 10^{16}\)
2. \(1.04\times 10^{13}\)
3. \(1.66\times 10^{16}\)
4. \(1.66\times 10^{15}\)
Subtopic:  Einstein's Photoelectric Equation |
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When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is \(\text 3.2~\text{V}.\) If a second light having wavelength twice of first light is used, the stopping potential drops to \(\text 0.7~\text{V}.\) The wavelength of first light is: (in m)
\(\left(h=6.63 \times 10^{-34} ~\text{J.s}, ~{e}=1.6 \times 10^{-19} ~\text{C}, {c}=3 \times 10^8 ~\text{m/s}\right)\)
1. \(\text 2.9 \times 10^{-8}\)
2. \(\text 2.2 \times 10^{-8}\)
3. \(\text 3.1 \times 10^{-7}\)
4. \(\text 2.5 \times 10^{-7}\)
Subtopic:  Einstein's Photoelectric Equation |
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Given below are two statements: 
Assertion (A): In photoelectric effect, on increasing the intensity of incident light the stopping potential increases.
Reason (R): Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency.
In the light of the above statements, choose the correct answer from the options given below:
1. Both (A) and (R) are True and (R) is the correct explanation of (A).
2. Both (A) and (R) are True but (R) is not the correct explanation of (A).
3. (A) is True but (R) is False.
4. (A) is False but (R) is True.
Subtopic:  Einstein's Photoelectric Equation |
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A monochromatic light is incident on a metallic plate having work function \(\phi\). An electron, emitted normally to the plate from a point \(A\) with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point \(B.\) The distance between \(A\) and \(B\) is: (Given: The magnitude of charge of an electron is \(e \) and mass is \(m, h \) is Planck's constant and \(c\) is velocity of light. Take the magnetic field exists throughout the path of electron)
1. \(\dfrac{\sqrt{2 {~m}({hc} / \lambda-\phi)}} { {eB} }\)

2. \(\dfrac{\sqrt{8 {~m}({hc} / \lambda-\phi)}} { {eB} }\)

3. \(\dfrac{\sqrt{{~m}({hc} / \lambda-\phi)}} { {eB} }\)

4. \(\dfrac{2\sqrt{{~m}({hc} / \lambda-\phi)}} { {eB} }\)
Subtopic:  Einstein's Photoelectric Equation |
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In the photoelectric effect, the stopping potential \(V_0\) v/s frequency \(v\) curve is plotted.
(\(h\) is Planck's constant and \(\phi_0\) is the work function of metal)
(A) \(V_0\) v/s \(v\) is linear
(B) The slope of \(V_0\) v/s \(v\) curve =\(\dfrac{\phi_0}{h}\)
(C) \(h\) constant is related to the slope of \(V_0\) v/s \(v\) line.
(D) The value of the electric charge of an electron is not required to determine h using the \(V_0\) v/s \(v\) curve.
(E) The work function can be estimated without knowing the value of \(h.\)
Choose the correct answer from the options given below -
1. (A), (C) and (E) only
2. (C) and (D) only
3. (A), (B) and (C) only
4. (D) and (E) only
 
Subtopic:  Einstein's Photoelectric Equation |
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