When electron with mass m is accelerated with volt V associated de-Broglie wavelength is `lambda` .When proton with mass M is accelerated with same voltage associated de-Broglie wavelength will be………

Options

A. `lambda((m)/(M))`
B. `lambda((M)/(m))`
C. `lambdasqrt((m)/(M))`
D. `lambdasqrt((M)/(m))`

Answer

C

Explanation

de-Broglie wavelength `lambda =(h)/(sqrt(2mVq))`
`therefore lambda prop (1)/(sqrt(M))`[`because` Remaining terms are same]
`therefore (lambda_(p))/(lambda_(e))=sqrt((m_(e))/(m_(p)))=sqrt((m)/(M))`
`therefore lambda_(p)=lambda sqrt((m)/(M)) [because lambda_(e)=lambda]`