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Question
The frequency and intensity of a light source are doubled. Consider the following statements.
(A) The saturation photocurrent remains almost the same.
(B) The maximum kinetic energy of the photoelectrons is doubled.
Options
A and B are true.
A is true but B is false.
A is false but B is true.
A and B are false.
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Solution
A is true but B is false.
Saturated current varies directly with the intensity of light. As the intensity of light is increased, a large number of photons fall on the metal surface. As a result, a large number of electrons interact with the photons. As a result, the number of emitted electrons increases and, hence, the current also increases.
At the same time, the frequency of the light source also increases.Also, with the increase in frequency of light, the stopping potential increases as well. This will reduce the current. The combined effect of these two is that the current will remain the same
Hence, A is true.
From the Einstein's photoelectric equation.
`K_max = hv - varphi`
Where `K_max` = kinetic energy of electron
v = frequency of light
`varphi` = work function of metal
It is clear from the above equation. As the frequency of light source is doubled, kinetic energy of electron increases. But, it becomes more than the double.
Hence, B is false.
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