If the momentum of electron is changed by `p_(m)` then the de-Broglie wavelength associated with it changes by 0.50%. The initial momentum of electron will be :

Options

A. `400p_(m)`
B. `200p_(m)`
C. `(p_(m))/(200)`
D. `(p_(m))/(400)`

Answer

A

Explanation

`p=(Deltap_(m))/(lambda)Deltal=(p_(m))/(200)"here" (Delta lambda)/(lambda)=(0*5)/(100)=(1)/(200)`