A deuteron and an alpha particl are accelerated with the same potential. Which one of the two has
greater value of de Broglie wavelength associated with it and i

Explanation

(i) Using de-Broglie wavelength formula, the dueteron and alpha particle are accelerated with same potential. So, boh their velocities are same.
`lambda = (h)/(sqrt(2 mV_(0) q)), lambda prop (1)/(sqrt(mq))`
So, `lambda_(due) prop (1)/(sqrt(2m_(d) q_(d))) and lambda_(alpha) prop (1)/(sqrt(8m_(alpha)q_(alpha)))`
`(lambda_(alpha))/(lambda_(d)) = (sqrt(2m_(d)q_(q)))/(sqrt(8m_(alpha) q_(alpha))) = sqrt((2)/(8)) = sqrt((1)/(4)) = (1)/(2)`
`lambda_(d) = 2 lambda_(alpha)`
(ii) For same potential of acceleration, KE is directly proportional to the .q.
Charge of duetron is +e
Charge of alpha is +2e
So, `K_(d) = (K_(alpha))/(2)`
Charge of alpha particle is more than the duetron.