Question

calculate de broglie wave length of an electron having kinetic energy 120eV​

Answer 1

Answer:

As p=2m K.E.

          =2×9×10−31×120×1.6×10−19

          =5.88×10−24kgms−1

de Broglie wavelength,

λ=ph=5.88×10−246.63×10−34=1.13×10−10m=1.13A0