Question

If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5\, p$ momentum of the photoelectron, the wavelength of the light should be:
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

• A. $\frac{1}{2}\lambda$
• B. $\frac{3}{4}\lambda$
• C. $\frac{2}{3}\lambda$
• D. $\frac{4}{9}\lambda$

Answer 1

Answer: (D)

$ hv -\phi= KE $
$ \Rightarrow\left(\frac{ hc }{\lambda}\right)_{\text {incident }}= KE +\phi $
$ \begin{array}{l} \left(\frac{ hc }{\lambda}\right)_{\text {incident }} \simeq KE \\ KE =\frac{ p ^{2}}{2 m }=\frac{ hc }{\lambda_{\text {incident }}}=\frac{ hc }{\lambda} \ldots(1) \end{array} $
$ \Rightarrow \frac{ p ^{2} \times(1.5)^{2}}{2 m }=\frac{ hc }{\lambda^{\prime}} $
divide (1) and (2)
$ (1.5)^{2}=\frac{\lambda}{\lambda^{\prime}} $
$ \Rightarrow \lambda^{\prime}=\frac{4 \lambda}{9} $