Q137 / 150 · MHT CET 2023 (09 May Shift 2) · Physics
When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for photoelectric current is $\mathrm{V}_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$, the stopping potential is $\frac{\mathrm{V}_0}{4}$, the threshold frequency of photoelectric emission is
- A$\frac{v}{6}$
- Correct answerB$\frac{v}{3}$
- C$\frac{2 v}{3}$
- D$\frac{4 v}{3}$
Solution
$\begin{aligned}
& \mathrm{eV}_0=h v-h v_0 ... (i)\\
& \frac{\mathrm{eV}_0}{4}=\frac{h v}{2}-h v_0 ... (ii)
\end{aligned}$
Dividing equation (i) by equation (ii),
$\begin{array}{ll}
& 4=\frac{v-v_0}{\frac{v}{2}-v_0} \\
\therefore & 2 v-4 v_0=v-v_0 \\
\therefore & 3 v_0=v \\
\therefore & v_0=\frac{v}{3}
\end{array}$
Hence, the correct answer is option: 'B'.