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.

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 :

Putting all given values in above equation, we get :

Hence, this is the required solution.

the answer is b ) square root of 5