Question

When a certain metal surface is illuminated with light of frequency ${}^{\prime }v{}^{\prime }$, 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)

$\frac{v}{3}$

Stopping potential is the energy required to stop the moving electrons after photoelectric emission. Hence it must be equal to the kinetic energy of the emitted electron.

Incident energy $=$ Threshold energy $+$ Kinetic energy

$⟹hv=h{v}_0+e{V}_0$ and $h\frac{v}{2}=h{v}_0+\frac{e{V}_0}{4}$

Solving the equations gives:

${\nu }_0=\frac{v}{3}$