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 :
[tex]\lambda = \dfrac{h}{\sqrt{2m(K.E)}}[/tex]
Putting all given values in above equation, we get :
[tex]\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[/tex]
Hence, this is the required solution.
