If p is the momentum of the fastest electron ejected from
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):

  1. $\frac{3}{4} \lambda$

  2. $\frac{1}{2} \lambda$

  3. $\frac{2}{3} \lambda$

  4. $\frac{4}{9} \lambda$


Correct Option: , 4

Solution:

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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