For photoelectric emission from certain metal the cut-off frequency is $v$. If radiation of frequency $2…
For photoelectric emission from certain metal the cut-off frequency is $v$. If radiation of frequency $2 \mathrm{v}$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $m$ is the electron mass)
$\sqrt{\frac{h v}{(2 m)}}$
$\sqrt{\frac{h v}{m}}$
$\sqrt{\frac{2 h v}{m}}$
$2 \sqrt{\frac{h v}{m}}$
Solution
$\begin{aligned}
& \text { } \mathrm{As} \frac{1}{2} m v_{\max }^2=h v \\
& \Rightarrow \quad v_{\max }^2=\frac{2 \mathrm{mN}}{m} \\
& \therefore \quad v_{\max }=\sqrt{\frac{2 h \mathrm{~V}}{\mathrm{~m}}} \\
&
\end{aligned}$