The threshold frequency of a metal is $f_{0}$. When the light of frequency $2 f_{0}$ is incident on the metal plate, the maximum velocity of photoelectrons is $v_{1}$. When the frequency of incident radiation is increased to $5 \mathrm{f}_{0}$, the maximum velocity of photoelectrons emitted is $v_{2}$. The ratio of $v_{1}$ to $v_{2}$ is :

Question

$\frac{v_{1}}{v_{2}}=\frac{1}{2}$

$\frac{v_{1}}{v_{2}}=\frac{1}{16}$

$\frac{v_{1}}{v_{2}}=\frac{1}{4}$

$\frac{v_{1}}{v_{2}}=\frac{1}{8}$

Solution

Please login to view the detailed solution steps...

Go to DASH