# The wavelength $\lambda _{e}$ of an electron and $\lambda _{p}$ of a photon are of same energy E are related by Option 1) $\lambda _{p}\alpha \lambda _{e}$ Option 2) $\lambda _{p}\alpha \sqrt{\lambda _{e}}$ Option 3) $\lambda _{p}\alpha \frac{1}{\sqrt{\lambda _{e}}}$ Option 4) $\lambda _{p}\alpha {\lambda _{e}}^{2}$

P Prateek Shrivastava

As we discussed in concept

Energy of a photon -

$\fn_jvn E= h\nu = \frac{hc}{\lambda }$

- wherein

$\fn_jvn h= Plank's\: constant$

$\boldsymbol{\nu= frequency\: of \: radiation }$

$\fn_jvn \lambda \rightarrow wave \: length$

&

De - Broglie wavelength with charged particle -

$\lambda = \frac{h}{\sqrt{2mE}}= \frac{h}{\sqrt{2mE}}$

$\lambda = \frac{h}{\sqrt{2mqv}}$

- wherein

$E\rightarrow kinetic\: energy\: o\! f particle$

$q\rightarrow charged \: particle$

For photon $E=\frac{hc}{\lambda p}$

For electron, $E=\frac{p^{2}}{2m}=\frac{h^{2}}{2m \lambda e^{2}}$

$\therefore\: \lambda p \propto \lambda e^{2}$

Option 1)

$\lambda _{p}\alpha \lambda _{e}$

This option is incorrect.

Option 2)

$\lambda _{p}\alpha \sqrt{\lambda _{e}}$

This option is incorrect.

Option 3)

$\lambda _{p}\alpha \frac{1}{\sqrt{\lambda _{e}}}$

This option is incorrect.

Option 4)

$\lambda _{p}\alpha {\lambda _{e}}^{2}$

This option is correct.

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