
The stopping potential for photoelectrons from a metal surface is $ {V_1} $ , when a monochromatic light of frequency $ {\nu _1} $ is incident on it. The stopping potential become $ {V_2} $ when monochromatic light of another frequency is incident on the same metal surface if $ h $ be the planck's constant and $ e $ be the change of an electron then the frequency of light in the second case is
(A) $ {\nu _1} - \dfrac{e}{h}({V_2} + {V_1}) $
(B) $ {\nu _1} + v\dfrac{e}{h}({V_2} + {V_1}) $
(C) $ {\nu _1} - \dfrac{e}{h}({V_2} - {V_1}) $
(D) $ {\nu _1} + \dfrac{e}{h}({V_2} - {V_1}) $
Answer
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Hint: To solve this problem we should know about the “Stopping potential”.
Stopping potential: it is the voltage difference required for stopping electrons from moving between plates and generating a current in the photoelectric experiment.
Monochromatic light: the light made up of a single component is called a Monochromatic light.
Complete step by step solution:
let, $ {\phi _0} $ is work function for the given metal
As given in the question,
When monochromatic light of frequency $ {\nu _1} $ is incident on it. Then stopping potential will be $ {V_1} $ .
So, $ h{\nu _1} = {\phi _0} + e{V_1} $ ……………….. $ (1) $
When monochromatic light of frequency \[{\nu _2}\] is incident on it. Then stopping potential will be $ {V_2} $ .
So, $ h{\nu _2} = {\phi _0} + e{V_2} $ ……………….. $ (2) $
From $ (1) $ and $ (2) $ . We get,
$ h\left( {{\nu _1} - {\nu _2}} \right) = e\left( {{V_1} - {V_2}} \right) $
$ \Rightarrow {\nu _1} - {\nu _2} = \dfrac{e}{h}\left( {{V_1} - {V_2}} \right) $
$ \Rightarrow {\nu _2} = {\nu _1} + \dfrac{e}{h}\left( {{V_2} - {V_1}} \right) $
Hence, frequency of second monochromatic light will be, $ {\nu _2} = {\nu _1} + \dfrac{e}{h}\left( {{V_2} - {V_1}} \right) $ .
So, option (d) is the correct option.
Note:
The photoelectric effect is emission of the electron when electromagnetic radiation, such as light, hits the surface of a material. Electrons emitted in this process are known as photoelectrons. Energy of an emitted electron depends on the frequency of light which falls on the surface of the material. This process is utilized in generating solar energy from solar panels by designing many photoelectric appliances like L.E.D, amplifiers etc.
Stopping potential: it is the voltage difference required for stopping electrons from moving between plates and generating a current in the photoelectric experiment.
Monochromatic light: the light made up of a single component is called a Monochromatic light.
Complete step by step solution:
let, $ {\phi _0} $ is work function for the given metal
As given in the question,
When monochromatic light of frequency $ {\nu _1} $ is incident on it. Then stopping potential will be $ {V_1} $ .
So, $ h{\nu _1} = {\phi _0} + e{V_1} $ ……………….. $ (1) $
When monochromatic light of frequency \[{\nu _2}\] is incident on it. Then stopping potential will be $ {V_2} $ .
So, $ h{\nu _2} = {\phi _0} + e{V_2} $ ……………….. $ (2) $
From $ (1) $ and $ (2) $ . We get,
$ h\left( {{\nu _1} - {\nu _2}} \right) = e\left( {{V_1} - {V_2}} \right) $
$ \Rightarrow {\nu _1} - {\nu _2} = \dfrac{e}{h}\left( {{V_1} - {V_2}} \right) $
$ \Rightarrow {\nu _2} = {\nu _1} + \dfrac{e}{h}\left( {{V_2} - {V_1}} \right) $
Hence, frequency of second monochromatic light will be, $ {\nu _2} = {\nu _1} + \dfrac{e}{h}\left( {{V_2} - {V_1}} \right) $ .
So, option (d) is the correct option.
Note:
The photoelectric effect is emission of the electron when electromagnetic radiation, such as light, hits the surface of a material. Electrons emitted in this process are known as photoelectrons. Energy of an emitted electron depends on the frequency of light which falls on the surface of the material. This process is utilized in generating solar energy from solar panels by designing many photoelectric appliances like L.E.D, amplifiers etc.
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