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When the light of frequency $2\nu_0$ (where $\nu_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$ . When the frequency of the incident radiation is increased to $5\nu_0$ , the maximum velocity of electrons emitted from the same plate is $v_2$ . The ratio of $v_1$ to $v_2$ is

4 : 1

1 : 4

1 : 2

2 : 1

Step-by-Step Solution

Using Einstein's photoelectric equation $E = W_0 + \frac{1}{2}mv^2$ . For frequency $2\nu_0$ , $h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2 \implies h\nu_0 = \frac{1}{2}mv_1^2$ (i). For frequency $5\nu_0$ , $h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2 \implies 4h\nu_0 = \frac{1}{2}mv_2^2$ (ii). Dividing (i) by (ii), we get $\frac{1}{4} = \frac{v_1^2}{v_2^2}$ , so $\frac{v_1}{v_2} = \frac{1}{2}$ .

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