Question

De-broglie postulated that the relationship,λ=h/p is valid for relativistic particles. Find out the de-broglie wavelength for an (relativistic) electron whose kinetic energy is 3MeV.​

Answer 1

Given:

De-broglie relation  $\lambda = \dfrac{h}{p}$

Kinetic energy = 3 MeV

To Find:

The de-broglie wavelength of the electron =?

Solution:

We know that,

$\lambda = \dfrac{h}{p}$   and  $E_{k} = \dfrac{1}{2}mv^{2}$

by using the above two equations, we can deduce a relation between de-broglie wavelength and kinetic energy, that can be shown as,

$\lambda = \dfrac{h}{\sqrt{2M_{e}E_{k} } }$

$\lambda = \dfrac{6.6 \times 10^{-34} }{\sqrt{2 \times 9.1 \times 10^{-31} \times 3\times 4.8 \times 10^{-3} } }$

on solving the above equation

$\lambda = \dfrac{6.6 \times 10^{-34} }{\sqrt{262.08 \times 10^{-34} } }$

$\lambda = \dfrac{6.6 \times 10^{-34} }{{16.18 \times 10^{-17} } }$

$\lambda = 4.08 \times 10^{-18}$m

Hence, the de-broglie wavelength is $4.08 \times 10^{-18}$m