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Light of frequency $v$ falls on material of threshold frequency $v_0$. Maximum kinetic energy of emitted electron is proportional to
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The correct answer is:
$v-v_0$
When light falls on a metallic surface, ejection of photoelectron results. In this process, conservation of energy holds.
Thus, from law of conservation of energy, Energy imparted by the photon
$=$ Maximum kinetic energy of the emitted electron + work function of the metal
or $\quad h v=(\mathrm{KE})_{\max }+\phi$
but $\phi=h v_0, v_0$ being threshold frequency.
$\begin{array}{ll}\therefore & (\mathrm{KE})_{\max }=h v-h v_0 \\ \text { or } & (\mathrm{KE})_{\max } \propto v-v_0\end{array}$
Thus, from law of conservation of energy, Energy imparted by the photon
$=$ Maximum kinetic energy of the emitted electron + work function of the metal
or $\quad h v=(\mathrm{KE})_{\max }+\phi$
but $\phi=h v_0, v_0$ being threshold frequency.
$\begin{array}{ll}\therefore & (\mathrm{KE})_{\max }=h v-h v_0 \\ \text { or } & (\mathrm{KE})_{\max } \propto v-v_0\end{array}$
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