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):
Correct Option: , 4
In photoelectric effect, $\frac{h c}{\lambda}=w+\mathrm{KE}$ of electron
Given that KE of ejected photoelectron is very high in comparison to work function $w$.
$\frac{h c}{\lambda}=\mathrm{KE} \Rightarrow \frac{h c}{\lambda}=\frac{\mathrm{P}^{2}}{2 m}$
New wavelength
$\frac{h c}{\lambda_{1}}=\frac{(1.5 \mathrm{P})^{2}}{2 m} \Rightarrow \lambda_{1}=\frac{4}{9} \lambda$
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