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

Answer

B

Explanation

`lambda=h/sqrt(2m qV)`
`lambda prop 1/sqrt(m)`
`lambda_(P)/lambda_(e)=sqrt(m_(e)/m_(P))`
`lambda_(P)=sqrt(m/M) lambda`