First slide

Photoelectric effect and study of photoelectric effect

Question

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

Moderate

A

double the earlier value
B

unchanged
C

more than doubled
D

less than doubled

Solution

$Let h v_{0} - W_{0} = K$
If frequency is doubled, let kinetic energy of photoelectrons be K₁.
$\begin{array}{l} 2 h v_{0} - W_{0} = K_{1} \\ \Rightarrow 2 (h v - W_{0}) + W_{0} = K_{1} \\ \Rightarrow 2 K + W_{0} = K_{1} \end{array}$
i.e., kinetic energy is more than doubled.

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