An electron of mass `m` when accelerated through a potential difference `V` has de - Broglie wavelength `lambda`. The de - Broglie wavelength associated with a proton of mass `M` accelerated through the same potential difference will be

Options

A. `lambda (M)/(m)`
B. `lambda(m)/(M)`
C. `lambda sqrt((M)/(m))`
D. `lambda sqrt((m)/(M))`

Answer

D

Explanation

`lambda = (h)/(sqrt(2mqV))`