Question 26

The threshold frequency of metal is $$f_0$$. When the light of frequency $$2f_0$$ is incident on the metal plate, the maximum velocity of photoelectron is $$v_1$$. When the frequency of incident radiation is increased to $$5f_0$$, the maximum velocity of photoelectrons emitted is $$v_2$$. The ratio of $$v_1$$ to $$v_2$$ is:

$$K_{\max} = \frac{1}{2}mv_{\max}^2 = h(f - f_0)$$

Given first condition ($$f = 2f_0$$, $$v_{\max} = v_1$$): $$\frac{1}{2}mv_1^2 = h(2f_0 - f_0) = hf_0$$

Given second condition ($$f = 5f_0$$, $$v_{\max} = v_2$$): $$\frac{1}{2}mv_2^2 = h(5f_0 - f_0) = 4hf_0$$

$$\frac{v_1^2}{v_2^2} = \frac{hf_0}{4hf_0} = \frac{1}{4} \implies \frac{v_1}{v_2} = \frac{1}{2}$$

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