MHT CET · Physics · Dual Nature of Matter
The frequency of incident light falling on a photosensitive material is doubled, the K.E. of the emitted photoelectrons will be
- A unchanged.
- B two times its initial value.
- C more than two times its initial value.
- D less than two times its initial value.
Answer & Solution
Correct Answer
(C) more than two times its initial value.
Step-by-step Solution
Detailed explanation
\(\mathrm{K} \cdot \mathrm{E}_{\max }=\mathrm{h} v-\mathrm{W}\)
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.
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.
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