When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for…

When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for photoelectric current is $\mathrm{V}_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$, the stopping potential is $\frac{\mathrm{V}_0}{4}$, the threshold frequency of photoelectric emission is
  1. $\frac{v}{6}$
  2. $\frac{v}{3}$
  3. $\frac{2 v}{3}$
  4. $\frac{4 v}{3}$

Solution

$\begin{aligned} & \mathrm{eV}_0=h v-h v_0 ... (i)\\ & \frac{\mathrm{eV}_0}{4}=\frac{h v}{2}-h v_0 ... (ii) \end{aligned}$ Dividing equation (i) by equation (ii), $\begin{array}{ll} & 4=\frac{v-v_0}{\frac{v}{2}-v_0} \\ \therefore & 2 v-4 v_0=v-v_0 \\ \therefore & 3 v_0=v \\ \therefore & v_0=\frac{v}{3} \end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

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