An electron of mass m has de broglie wavelength `lamda` when accelerated through a potential difference V . When a proton of mass M is accelerated through a potential difference 9V, the de broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage ) .

Options

A. `(lamda)/(3) sqrt((M)/(m))`
B. `(lamda)/(3)(M)/(m)`
C. `(lamda)/(3)sqrt((m)/(M))`
D. `(lamda)/(3)(m)/(M)`

Answer

C

Explanation

According to question,
Mass of electron=m
Wavelength of electron`=lamda`
Potential difference in case of electron=V
From de-Broglie relation,
`lamda=(h)/(p) implies lamda=(h)/(sqrt(2mKE))=(h)/(sqrt(2mqV))`
`implies lamda prop (1)/(sqrt(qVm))`
For electron `lamda prop (1)/(sqrt(eVm))`
For proton, `lamda_(1)=(1)/(sqrt(e9VM))`
where, e is the charge on proton, potential difference =9V mass of proton=m
From eqs. (i) and (ii), we get
`(lamda)/(lamda_(1))=sqrt((9VMe)/(eVm)) implies lamda_(1)=(lamda)/(3)sqrt((m)/(M))`