A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is

Options

A. `sqrt(2):1`
B. `sqrt(4):1`
C. `sqrt(6):1`
D. `sqrt(8):1`

Answer

D

Explanation

(d) : As `lamda=(h)/(sqrt(2mqV))`
`:.lamdaprop(1)/(sqrt(mq)):.(lamda_(p))/(lamda_(alpha))=(sqrt(m_(alpha)q_(alpha)))/(sqrt(m_(p)q_(p)))`
`(sqrt(4m_(p)xx2e))/(sqrt(m_(p)xxe))=sqrt(8)" "( :.m_(alpha)=4m_(p),q_(alpha)=2q_(p))`