An electron of mass m has de-Broglie wavelength ·A when accelerated through potential differe_nce V. When proton of mass M, is accelerated through potential diff ere nee 9 V,. the de-BrOglie wavelength associated v,rith it will be_ (Assume -that wavelength is-determin•ed. at low voltage)

Options

A. `lambda/3sqrtM/m`
B. `lambda/3M/m`
C. `lambda/3sqrtm/M`
D. `lambda/3m/M`

Answer

C

Explanation

According to question, Mass of electron = m brgt Wavelength of electron = `lambda`
Potential difference in case of electron = V
From de-Broglie relation,
`lambda= h/p Rightarrow lambda-h/(sqrt2mKE)=h/(sqrt2mqV)`
`Rightarrow lambda prop 1/(sqrtqVm)`
For electron `lambda prop 1/(sqrteVm)`
where, e is the charge on proton, potential difference = 9V
Mass of proton = m
From Eqs. (i) and (ii), we get
`lambda/lambda_1=sqrt((9VMe)/(eVm))=Rightarrow lambda_1=lambda/3sqrt(m/M`