The threshold frequency of a metal is ' $F_0$ '. When light of frequency $2 \mathrm{~F}_0$ is incident on…

The threshold frequency of a metal is ' $F_0$ '. When light of frequency $2 \mathrm{~F}_0$ is incident on the metal plate, the maximum velocity of photoelectron is ' $\mathrm{V}_1$ ' When the frequency of incident radiation is increased to ' $5 \mathrm{~F}_0$ ', the maximum velocity of photoelectrons emitted is ' $\mathrm{V}_2$ '. The ratio of $\mathrm{V}_1$ to $\mathrm{V}_2$ is
  1. $\frac{1}{8}$
  2. $\frac{1}{16}$
  3. $\frac{1}{4}$
  4. $\frac{1}{2}$

Solution

$\begin{array}{ll} & K E_{\max }=h F-F_0 \\ & \text { When, } \mathrm{F}=2 \mathrm{~F}_0 \\ & \frac{1}{2} m V_1^2=2 \mathrm{hF}_0-\mathrm{F}_0=\mathrm{F}_0 \\ & \text { When, } \mathrm{F}=5 \mathrm{~F}_0 \\ & \frac{1}{2} m V_2^2=5 \mathrm{hF}_0-\mathrm{F}_0=4 \mathrm{~F}_0 \\ \therefore \quad & \left(\frac{\mathrm{~V}_1}{\mathrm{~V}_2}\right)^2=\frac{\mathrm{F}_0}{4 \mathrm{~F}_0} \\ \therefore \quad & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\frac{1}{2}\end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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