Question

Proton, deuteron and alpha particles of the same kinetic energy and moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particles are respectively ${{r}_{p}}$, ${{r}_{d}}$ and${{r}_{\alpha }}$. Which one of the following relations is correct?
A. ${{r}_{\alpha }}={{r}_{p}}={{r}_{d}}$
B. ${{r}_{\alpha }}={{r}_{p}}\langle {{r}_{d}}$
C. ${{r}_{\alpha }}\rangle {{r}_{d}}\rangle {{r}_{p}}$
D. ${{r}_{\alpha }}={{r}_{d}}\rangle {{r}_{p}}$

Answer 1

Hint: Remember that the particle stays in its circular path due to the centripetal force acting on it. This force is provided by the magnetic force and you could equate both to get the radius of the circular path of each particle. Now you could substitute for momentum in terms of kinetic energy and then do a comparison of the radii using only the non constant terms. Try to express the mass and charge of the deuteron and alpha particle in terms of that of the proton and hence do the comparison.

Formula used:
Expression for centripetal force,
${{F}_{C}}=\dfrac{m{{v}^{2}}}{r}$
Expression for magnetic force,
${{F}_{B}}=qv\times B=qvB\sin \theta $
Expression for kinetic energy,
$K.E=\dfrac{{{P}^{2}}}{2m}$

Complete answer:
We are given 3 particles, namely, proton, deuteron and alpha particles having the same kinetic energy and they are moving in circular paths in a constant magnetic field. Using these conditions, we are asked to find the relation between the radii of the circular trajectories of each particle.
We know that, for a particle moving around in a circular path with radius r there is a force acting perpendicular to the path towards the centre of the circle called the centripetal force given by the expression,
${{F}_{C}}=\dfrac{m{{v}^{2}}}{r}$ ………………………………. (1)
We know that the magnetic force is given by the expression,
${{F}_{B}}=qv\times B=qvB\sin \theta $
But for magnetic field perpendicular to the velocity, θ=90°
$\Rightarrow {{F}_{B}}=qvB$ ………………………… (2)
We also know that the magnetic force is responsible for the circular motion and hence the centripetal force. Therefore, we could equate (1) and (2),
$\Rightarrow qvB=\dfrac{m{{v}^{2}}}{r}$
$\Rightarrow r=\dfrac{mv}{qB}$
But $P=mv$
$\Rightarrow r=\dfrac{P}{qB}$ …………………… (3)
But we know the expression for kinetic energy given by,
$K.E=\dfrac{{{P}^{2}}}{2m}$
$\Rightarrow P=\sqrt{2mK.E}$
Substituting this in equation (3) we get,
$r=\dfrac{\sqrt{2mK.E}}{qB}$
We are given in the question that the K.E and magnetic field is same, so,
$r\propto \dfrac{\sqrt{m}}{q}$ …………………………. (4)
Let mass of proton be ${{m}_{p}}$ then, mass of deuteron is twice that of proton and also the mass of alpha particle is four times that of proton, that is,
${{m}_{d}}=2{{m}_{p}}$
${{m}_{\alpha }}=4{{m}_{p}}$
We know that, charge of electron and proton is same, e, then, the charge of deuteron is also same e and the charge of alpha particle is 2e, that is,
${{q}_{p}}={{q}_{d}}=e$
${{q}_{\alpha }}=2e$
Using these values in (4), if the radii of proton, deuteron and alpha particle are given by${{r}_{p}}$, ${{r}_{d}}$ and ${{r}_{\alpha }}$ respectively then, their ratios are given by,
${{r}_{p}}:{{r}_{d}}:{{r}_{\alpha }}=\dfrac{\sqrt{{{m}_{p}}}}{e}:\dfrac{\sqrt{2{{m}_{p}}}}{e}:\dfrac{\sqrt{4{{m}_{p}}}}{2e}$
$\Rightarrow {{r}_{p}}:{{r}_{d}}:{{r}_{\alpha }}=1:\sqrt{2}:1$
So clearly, ${{r}_{p}}$ and ${{r}_{\alpha }}$ are equal and are less than${{r}_{d}}$.
Therefore, ${{r}_{\alpha }}={{r}_{p}}\langle {{r}_{d}}$

Hence, the answer to the question is option B.

Note:
For problems like this which involves comparison of quantities, it is better that you go for direct comparison by using the given quantities. While doing the comparison make sure that you include all the quantities except for the constant ones. Also, you should know the masses and charges of proton, deuteron and alpha particles, not necessarily their values but you should be at least able to compare them.