When a certain metal surface is illuminated with light of frequency v, the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$ the stopping potential is $\frac{V_0}{4}$. The threshold frequency for photoelectric emission is:

$\frac{v}{3}$

According to Einstein’s photoelectric equation $K_{max} = hv − \phi_0$

Where v is the incident frequency and $\phi_0$ is the work function of the metal.

As $K_{max} = eV_0$

Where V₀ is the stopping potential. Therefore

$eV_0=hv-\phi_0$  (i)

and $e\frac{V_0}{4}=h\frac{v}{2}-\phi_0$   (ii)

From (i) and (ii), we get

$\frac{hv}{4}-\frac{\phi_0}{4}=\frac{hv}{2}-\phi_0$ or $\phi_0-\frac{\phi_0}{4}=\frac{hv}{2}-\frac{hv}{4}$

$\frac{3}{4}\phi_0=\frac{hv}{4}$ or $\phi_0=\frac{hv}{3}$

∴ Threshold frequency, $v_0=\frac{\phi_0}{h}=\frac{hv}{3}×\frac{1}{h}=\frac{v}{3}$