JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 15)
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 :
$$\frac{v_{1}}{v_{2}}=\frac{1}{2}$$
$$\frac{v_{1}}{v_{2}}=\frac{1}{16}$$
$$\frac{v_{1}}{v_{2}}=\frac{1}{4}$$
$$\frac{v_{1}}{v_{2}}=\frac{1}{8}$$
Explanation
$$
\begin{aligned}
& \frac{1}{2} m v^2=h f-h f_0 \\\\
& \Rightarrow \frac{1}{2} m v_1^2=2 h f_0-h f_0=h f_0 \\\\
& \text { also, } \frac{1}{2} m v_2^2=5 h f_0-h f_0=4 h f_0
\end{aligned}
$$
taking ratio,
$$ \frac{v_1^2}{v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1}{v_2}=\frac{1}{2} $$
taking ratio,
$$ \frac{v_1^2}{v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1}{v_2}=\frac{1}{2} $$
Comments (0)
