Question

An `alpha`-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelength associated with them.

Answer 1

Correct Answer - `1//(2sqrt2)`
The kinetic energy gained by a charged particle of charge q, mass m, moving with velocity v, when acceleration through the potential difference V is K.E. `=1/2mv^(2)=qV or mv=[2mqV]^(1//2)`
De-Broglie wavelength of the particle is
`lambda=h/(mv)=h/([2mqV]^(1//2))`
so `lambda prop 1/(sqrt(mq))` (for the same value of V)
`:. (lambda_(alpha))/(lambda_(p))=sqrt((m_(p)q_(p))/(m_(alpha)q_(alpha)))=sqrt((m_(p)xxe)/(4m_(p)xx2e)) =1/(sqrt(8)) =1/(2sqrt(2))`