A proton and alpha particle are accelerated through the same accelerating potential. Which one of the two has (a) greater value of de-broglie wavelength associated with it, and (b) less kinetic energy? justify your answer.

Explanation

(a) De-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that `lambda prop 1` (Mass) (charge).
Mass and charge of a proton are mp and e respectively and, mass and charge of an alpha particle are `4m^(rho)` and 2e respectively. Where, e is the charge of an electron
`lambda_(rho)lambda_(alpha)=M_(a)q_(a)M_(p)q_(p)=malphaq=(4m_(p))(2e)(m_(p))(e)=22`
Thus, d-Broglie wavelength associated with proton is 22 times of the de-Broglie wavelength of alpha particle.
(b) K. E `prop q` (For same accelerating potential ). Charge of an alpha particle is more as compared to a proton. So, it will have a greater value of K.E. Hence, proton will have lesser kinetic energy.