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 :
Solution
$$
\begin{aligned}
& \frac{1}{2} m v^2=h f-h f_0 \\\\
& \Rightarrow \frac{1}{2} m v_1^2=2 h f_0-h f_0=h f_0 \\\\
& \text { also, } \frac{1}{2} m v_2^2=5 h f_0-h f_0=4 h f_0
\end{aligned}
$$
<br/><br/>taking ratio,
<br/><br/>$\frac{v_1^2}{v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1}{v_2}=\frac{1}{2}$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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