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. `lamda(m)/M`
B. `lamdasqrt(m/M)`
C. `lamda(M)/m`
D. `lamdasqrt(M/m)`

Answer

B

Explanation

`lamda=h/(sqrt(2mqV))implies lamdaprop1/sqrtm`
`lamda_e/lamda_p=sqrt(m_p/m_e)implies lamda_p = lamda sqrt(m/M)`