An electron and a proton are accelerated through the same potential. Which one of the two has (i)b greater value of de-Broglie wavelength associated with it and (ii) less momentum?justify your answer.

Explanation

As a charge q is accelerated by a potential V, we have `W = qV = (1)/(2) mv^(2) = (p^(2))/(2m)`
As the work done on the charge become K.E., `p = sqrt(2mq V)`
(i) Since `lamda = (h)/(p') (lamda_(e))/(lamda_(p)) = sqrt((2 m_(e) eV)/(2m_(p) eV)) = sqrt((m_(e))/(m_(p)) lt 1`
So the electron will have greater de-Broglie wavelength
(ii) Since `p prop sqrtm and m_(e) lt m_(p)` momentum is less iwht electron.