Question 17

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:

Using Einstein's photoelectric equation: $$KE_{max} = h\nu - h\nu_0$$

For the first case, $$\nu = 2f_0$$

$$ \frac{1}{2}mv_1^2 = h(2f_0) - hf_0 = hf_0 $$

For the second case, $$\nu = 5f_0$$

$$ \frac{1}{2}mv_2^2 = h(5f_0) - hf_0 = 4hf_0 $$

Taking the ratio:

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

Therefore, $$v_1 : v_2 = 1 : 2$$.

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