Question

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 reason to justify your answer.

Answer 1

From law of conservation of energy, Kinetic energy equals the change in electrical energy in the particle.

$\frac{1}{2}m{v}^2=qV$

$p=mv=\sqrt{2mqV}$

De-broglie wavelength of a particle is given by: $\lambda =\frac{h}{p}$

$\lambda =\frac{h}{\sqrt{2mqV}}$

(a) Ratio of de-broglie wavelength of proton to deuteron is given by:

$\frac{{\lambda }_p}{{\lambda }_d}=\frac{\sqrt{m_d{q}_d}}{\sqrt{m_p{q}_p}}=\sqrt{\frac{2\times 1}{1\times 1}}=\sqrt{2}$

Hence, proton has a higher de-broglie wavelength.

(b) $p=\frac{h}{\lambda }$

Hence, deuteron has higher momentum.