The threshold frequency of a metal is $f_0$. When the light of frequency $2 f_0$ is incident on the

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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 $5 f_0$, the maximum velocity of photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is

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(d): Threshold frequency $=f_0$ According to the Einstein's photoelectric equation, $\mathrm{KE}_{	ext {max }}=E-\phi$ \ Case I: $\frac{1}{2} m v_1^2=h 	imes 2 f_0-h f_0 \quad$ or $\quad \frac{1}{2} m v_1^2=h f_0$ \ Case II : $\frac{1}{2} m v_2^2=h 	imes 5 f_0-h f_0 \quad$ or $\quad \frac{1}{2} m v_2^2=4 h f_0$ From case I and case II, $$ \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1^2}{v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1}{v_2}=\frac{1}{2} $$