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. `lambdasqrt(M/m)`
B. `lambdasqrt(m/M)`
C. `lambda(M/m)`
D. `lambda(m/M)`

Answer

B

Explanation

`lambda=h/(sqrt(2mK))rArr (lambda_(2))/(lambda_(1))=sqrt(M/m)`
K=qV is same for both proton and electrons.