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The maximum velocity of electrons emitted from. a metal surface is $\mathrm{v}$, when frequency of light falling on it is $f$. The maximum velocity when frequency becomes $4 \mathrm{f}$ is
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Verified Answer
The correct answer is:
$>2 \mathrm{v}$
According to Einstein's photoelectric equation,
$$
E=W_{0}+\frac{1}{2} m v_{\max }^{2} \Rightarrow v_{\max }=\sqrt{\frac{2\left(h f-W_{0}\right)}{m}}
$$
If frequency becomes $4 f$ then
$$
\begin{array}{l}
v^{\prime}=\sqrt{\frac{2\left(h \times 4 f-W_{0}\right)}{m}}=2 \sqrt{\frac{2\left(h f-\frac{W_{0}}{4}\right)}{m}} \\
\Rightarrow v^{\prime}>2 v
\end{array}
$$
$$
E=W_{0}+\frac{1}{2} m v_{\max }^{2} \Rightarrow v_{\max }=\sqrt{\frac{2\left(h f-W_{0}\right)}{m}}
$$
If frequency becomes $4 f$ then
$$
\begin{array}{l}
v^{\prime}=\sqrt{\frac{2\left(h \times 4 f-W_{0}\right)}{m}}=2 \sqrt{\frac{2\left(h f-\frac{W_{0}}{4}\right)}{m}} \\
\Rightarrow v^{\prime}>2 v
\end{array}
$$
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