An electron of mass 'm', when accelerated through a potential 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. `sqrt(m/M)lamda`
B. `sqrt(M/m)lamda`
C. `(lamdam)/(m)`
D. `(M)/(lamdam)`

Answer

A

Explanation

Both have the same charged hence they acquire the same K.E.
`lamda = (h)/(sqrt(2mE)) " for same " E, lamda prop 1/sqrtm`
`therefore lamda_P/lamda_e = sqrt(m/M) therefore lamda_P = sqrt(m/M) lamda_e " or " sqrt(m/M) lamda`