Question

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

Answer 1

Given :

An electron with kinetic energy of 3 MeV.

To Find :

The de-Broglie wavelength for that electron.

Solution :

We know, de-Broglie wavelength for an electron with kinetic energy K.E is given by :

$\lambda = \dfrac{h}{\sqrt{2m(K.E)}}$

Putting all given values in above equation, we get :

$\lambda = \dfrac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 3\times 10^6 \times 1.6\times 10^{-19}}}\\\\\\\lambda = 7.089 \times 10^{-13} \ m$

Hence, this is the required solution.