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:
Answer & explanation
Correct answer: option 2
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, $v_0=\frac{\phi_0}{h}=\frac{hv}{3}×\frac{1}{h}=\frac{v}{3}$