The proton when accelerated through a potential difference of V volt has a wavelength `lamda` associated with it. If an `alpha`-particle is to have the same wavelength `lamda`, it must be accelerated through a potential difference of

Options

A. `(V)/(8)`
B. `(V)/(4)`
C. 4V
D. 8V

Answer

A

Explanation

When a particle of charge q and mass m is accelerated across a potential difference of V, its wavelength is given by
`lamda = (h)/(sqrt(2mqV)) or, V =(h^(2))/(2mq lamda^(2))`
For proton, `V_(p) = (h^(2))/(2m_(p) q_(p) lamda^(2))`
For `alpha`-particle, `V_(alpha) = (h^(2))/(2m_(alpha) q_(alpha) lamda^(2))`
or, `(V_(alpha))/(V_(p)) = (m_(p))/(m_(alpha)) xx (q_(p))/(q_(alpha)) = (1)/(4) xx (1)/(2) = (1)/(8)`