A proton and an `alpha` particle are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths is

Options

A. 2
B. `sqrt8`
C. `(1)/(sqrt8)`
D. 1

Answer

B

Explanation

Kinetic energy of charged, particle, accelerated through a potential difference v,
i.e., `K=qV`
`:.` Momentum of the particle,
`p=sqrt(2mK)= sqrt(2mqV)`
de-Broglie wavelenth of the particle
`lambda=(h)/(p)=(h)/(sqrt(2mqV))`
`:.m_(alpha)=4m_(p) and q_(2)=2q_(1)`
`:. (lambda_(p))/(lambda_(alpha))=sqrt((m_(2)q_(2))/(m_(1)q_(1)))= sqrt((4xx2)/(1xx1))= sqrt(8)`