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

Option: 1

3/4\lambda


Option: 2

1/2\lambda


Option: 3

2/3\lambda


Option: 4

4/9\lambda


Answers (1)

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As we learnt,

\begin{array}{l}{\mathrm{E}=\phi+\mathrm{KE}} \\ {\mathrm{E}=\mathrm{KE}} \\ {\frac{\mathrm{h}c}{\lambda}=\frac{1}{2} \mathrm{mv}^{2}\left(\frac{\mathrm{m}}{\mathrm{m}}\right)=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}} \\ {\mathrm{p}^{2} \propto \frac{1}{\lambda}}\end{array}


Now we have:
\begin{array}{l}{\left(\frac{P_{2}}{P_{1}}\right)^{2}=\frac{\lambda_{1}}{\lambda_{2}}} \\ {\left(\frac{1.5 P_{1}}{P_{1}}\right)^{2}=\frac{\lambda_{1}}{\lambda_{2}}} \\ {\left(\frac{3}{2}\right)^{2}=\frac{\lambda_{1}}{\lambda_{2}}} \\ {\frac{9}{4}=\frac{\lambda_{1}}{\lambda_{2}}} \\ {\lambda_{2}=\frac{4}{9} \lambda_{1}}\end{array}

Therefore, Option(4) is correct.

Posted by

Kuldeep Maurya

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