The frequency of incident light falling on a photosensitive material is doubled, the K.E. of the emitted…

The frequency of incident light falling on a photosensitive material is doubled, the K.E. of the emitted photoelectrons will be
  1. unchanged.
  2. two times its initial value.
  3. more than two times its initial value.
  4. less than two times its initial value.

Solution

$\mathrm{K} \cdot \mathrm{E}_{\max }=\mathrm{h} v-\mathrm{W}$
When frequency is doubled, $\begin{aligned} K \cdot E_{\max } & =\mathrm{h} v-\mathrm{W} \\ & =2 \mathrm{~h} v-2 \mathrm{~W}+\mathrm{W} \\ & =2(\mathrm{~h} v-\mathrm{W})+\mathrm{W} \\ & =2 \mathrm{~K}+\mathrm{W} \end{aligned}$ $\therefore \quad$ The K.E. $\max$ is more than twice the initial value.

Asked in: MHT CET 2024 (03 May Shift 2)

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