A proton and a deuteron 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 momentum?
Give reasons to justify your answer.De-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that
Wavelength is inversely proportional to q and m.
Now, for proton and deuteron, we have 
where, e is the charge of an electron and mp is the mass of proton.
Thus, de-Broglie wavelength associated with proton is 2√ times of the de-Broglie wavelength of deuteron and hence it is more.
(b) Momentum is inversely proportional to the wavelength.
Mathematically it is given by,
;
where, h = plank's constant.
Wavelength of a proton is more than that of deuteron thus the momentum of a proton is lesser than that of deuteron.
