In a photoemissive cell with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to $\frac{λ}{4}$, the speed of the fastest emitted electron will be:
Answer & explanation
Correct answer: option 4
As, $\frac{1}{2}mv_{max}^2=\frac{hc}{λ}-\phi_0$
$m_{max}=\sqrt{\frac{2}{m}(\frac{hc}{λ}-\phi_0)}=v$
$v'_{max}=\sqrt{\frac{2}{m}(\frac{hc}{λ/4}-\phi_0)}$
$=2\sqrt{\frac{2}{m}(\frac{hc}{λ}-\phi_0)+\frac{6}{m}\phi_0}$
> 2v