Q154 / 160 · WBJEE 2012 · Physics
When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for photoelectric current is $V_{0}$. When the same surface is illuminated by light of frequency $\frac{v}{2}$ the stopping potential is $\frac{V_{0}}{4} .$ The threshold
frequency for photoelectric, emission is
- A$\frac{v}{6}$
- Correct answerB$\frac{v}{3}$
- C$\frac{2 v}{3}$
- D$\frac{4 v}{3}$
Solution
Ist case
$$
\begin{array}{l}
h v=h v_{0}+e V_{0} \\
\frac{h v}{2}=h v_{0}+\frac{e V_{0}}{4}
\end{array}
$$
Solving Eqs. (i) and (ii), we get
$$
v_{0}=\frac{v}{3}
$$
Hence, the correct answer is option: 'B'.