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

(i) The de-Broglie wavelength `lamda` associated with same potential V is
`lamda = (h)/(sqrt(2 meV)) " " :. lamda alpha (1)/(sqrtm)` [For same potential]
Since mass of alpha particle is larger than a proton
`:. lamda_("proton") gt lamda_("alpha")`
(ii) Since, `KE = hv = h (c)/(lamda)`
`E alpha (1)/(lamda)`
Since `lamda_("proton") gt lamda_("alpha")`
`:. K_("proton")` is less than `K_("alpha")`