An electron of mass m, when accelerated through a potential difference V, has de-Broglie wavelength `lamda`, the de-Broglie wavelength associated with a protom of mass M and accelerated through the same potential difference V will be

Options

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

Answer

B

Explanation

As per relation: `lamda=(h)/(sqrt(2mqV))` for same potential V, we have
`(lamda_(p))/(lamda_(c))=(lamda_(p))/(lamda)=sqrt((m)/(M)) or lamda_(p)=lamdasqrt((m)/(M))`.