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 = (h)/(√(2m(K.E)))`

Putting all given values in above equation, we get :

`\lambda = \frac{6.626* 10^(-34)}{\sqrt{2* 9.1* 10^(-31)* 3* 10^6 * 1.6* 10^(-19)}}\\\\\\\lambda = 7.089 * 10^(-13) \ m`

Hence, this is the required solution.