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

Answer

B

Explanation

Momentum of electron , `P_e =sqrt(2meV)`
Momentum of proton , `p_p = sqrt(2MeV)`
`:. lamda_p/lamda_e =(h//P_p)/(h//P_e)=p_e/p_p=(sqrt(2meV))/(sqrt(2MeV))=sqrt(m/M)`
`:. lamda_p = lamda_e sqrt(m/M)=lamda sqrt(m/M)" " ` (Given `lamda_(e) = lamda` )