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}

 

Answers (1)
P Prateek Shrivastava

As we discussed in concept

Energy of a photon -

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

- wherein

h= Plank's\: constant

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

\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
ightarrow kinetic: energy: o! f particle

q
ightarrow 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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