Two paricles have idential charges. If they are accelerated throgh identical potential differences, then the ratio of their deBrogile wavelength would be

Options

A. `lambda_(1) : lambda_(2) = 1 : 1`
B. `lambda_(1) : lambda_(2) = m_(2) : m_(1)`
C. `lambda_(1) : lambda_(2) = sqrt(m_(2)) : sqrt(m_(1))`
D. `lambda_(1) : lambda_(2) = sqrt(m_(1)) : sqrt(m_(2))`

Answer

C

Explanation

`K.E. = qDeltaV`
`lambda_(1) = (h)/(P_(1)) = (h)/(sqrt(2m_(1)K_(1))) : lambda_(2) - (h)/(p_(2)) = (h)/(sqrt(2)m_(2)K_(2))`
`:' q = q_(1) = q_(2)` & `DeltaV` is same `:. (lambda_(1))/(lambda_(2)) = sqrt((m_(2))/(m_(1))`