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I vs V - Varying Frequency

The stopping ...

Question

The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_{1}$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ be the Plancks onstant and $e$ be the charge of an electron, then the frequency of light in the second case is

A

v_{1} - \frac{e}{h} (V_{2} + V_{1})

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B

v_{1} + \frac{e}{h} (V_{2} + V_{1})

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C

v_{1} - \frac{e}{h} (V_{2} - V_{1})

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D

v_{1} + \frac{e}{h} (V_{2} - V_{1})

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Solution

The correct option is C $v_{1} + \frac{e}{h} (V_{2} - V_{1})$
Let the work function of the metal be $ϕ$ .
Stopping potential of the metal is equal to the maximum kinetic energy of the photoelectrons emitted.
$∴$ $e V = K . E_{m a x} = h v - ϕ$
For frequency $v_{1}$ :
$e V_{1} = h v_{1} - ϕ$ .......(1)
For frequency $v_{2}$ :
$e V_{2} = h v_{2} - ϕ$ .......(2)
From equation (2) - (1) we get, $e (V_{2} - V_{1}) = h v_{2} - h v_{1}$
$⟹$ $v_{2} = v_{1} + \frac{e}{h} (V_{2} - V_{1})$

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