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

Option: 1
$\frac{\nu}{6}$

Option: 2
$\frac{\nu}{3}$

Option: 3
$\frac{2 \nu}{3}$

Option: 4
$\frac{4 \nu}{3}$

Answers (1)

best_answer

According to Einstein's photoelectric equation $\mathrm{K}_{\max }=\mathrm{h\nu}-\phi_0$

Where $\mathrm{\nu}$ is the incident frequency and $\phi_0$ is the work function of the metal.

As $\mathrm{K}_{\max }=\mathrm{eV}_0$

Where $\mathrm{V}_0$ is the stopping potential.

$\therefore \mathrm{eV}_0=\mathrm{h\nu}-\phi_0 \quad \quad \quad \quad \quad \dots(i)$

and $\mathrm{e} \frac{\mathrm{V}_0}{4}=\mathrm{h} \frac{\mathrm{\nu}}{2}-\phi_0 \quad \quad \quad \quad \quad \dots(ii)$

From (i) and (ii), we get

$\frac{\mathrm{h\nu}}{4}-\frac{\phi_0}{4}=\frac{\mathrm{h\nu}}{2}-\phi_0 \quad \text{ or }\ \ \ \phi_0-\frac{\phi_0}{4}=\frac{h \nu}{2}-\frac{h \nu}{4}$
$\frac{3}{4} \phi_0=\frac{\mathrm{h\nu}}{4} \quad \text{ or }\quad \phi_0=\frac{\mathrm{h\nu}}{3}$

$\therefore$ Threshold frequency,

$\mathrm{\nu}_0=\frac{\phi_0}{\mathrm{~h}}=\frac{\mathrm{h\nu}}{3} \times \frac{1}{\mathrm{~h}}=\frac{\mathrm{\nu}}{3}$

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