Question

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. $\frac{v}{6}$
• B. $\frac{v}{3}$
• C. $\frac{2v}{3}$
• D. $\frac{4v}{3}$

Answer 1

Answer: (B)

Ist case
$h v=h v_{0}+e V_{0}$ ...(i)
$\frac{h v}{2}=h v_{0}+\frac{e V_{0}}{4}$ ...(ii)
Solving Eqs. (i) and (ii), we get
$V_{0}=\frac{v}{3}$