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 wavelenth of a particle depends upon its mass and chargbe for same accelerating potential, such that
`lambda prop (1)/(sqrt("mass"xx "charge")`
Mass and charge of a deuteron are `2m_(rho)and e` respectively and mass and charge of an alpha particle are `4m^(rho)` and 2e respectively. Where, `m_(rho)` is the mass of a proton and e is the charge of an electron.
`(lambda_(d))/(lambda_(prop))=(sqrt(4mp.2e))/(sqrt(2mp.e))=2`
`rArr" "lambda_(d)=2lambda_(prop)`.