Question

A proton and a deuteron are accelerated through the same acceleration 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

Hint: In the given question, the base concept is the conservation of energy of particles. Due to which the particles have a momentum and have De-Broglie wavelengths associated with them. We can thus find the relation between De-Broglie wavelength, momentum and Planck’s constant to solve the question further.
Formula used: $\lambda ={\displaystyle \frac{h}{p}}$

Complete answer:
a).The De-Broglie wavelength of a particle is:
$\lambda ={\displaystyle \frac{h}{p}}$
Here, $h$ is the Planck’s constant.
$p$ is the momentum of the particle and its value is:
[\p=\sqrt{2mqV}\]
Here $m$ is the mass of the particle, $q$ is the charge on the particle and $V$ is the velocity of the particle.
Since proton and deuteron are accelerated through the same acceleration potential hence their velocities will also be equal and both of them will have the same charge. Thus, we can conclude that:
$$\begin{array}{rl}& \lambda \propto {\displaystyle \frac{1}{p}}\\ & \Rightarrow \lambda \propto {\displaystyle \frac{1}{\sqrt{m}}}\end{array}$$
Mass of the deuteron is twice the mass of the proton, hence the proton will have a greater value of De-Broglie wavelength associated with it.
b).The momentum of a particle is given by:
$p={\displaystyle \frac{h}{\lambda }}$
Here, the momentum of the particle is inversely proportional to the De-Broglie wavelength of the particle, i.e.,
$p\propto {\displaystyle \frac{1}{\lambda }}$
Since, proton has a greater value of De-Broglie wavelength associated with it, hence it will have less momentum as compared to the momentum of deuteron.

Note:
Proton and deuteron, both are the isotopes of hydrogen and thus all of them have the same charge, i.e., same atomic number but different masses, i.e., different mass number due to which we observe a difference in momentum of the particles and De-Broglie wavelength associated with the particles.