Question

A proton, a deuteron, and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is

Answer 1

Step 1: Equation force due to magnetic field and force due to centripetal acceleration

The force exerted by a uniform magnetic field on a charged particle is given as,
$F=qvB\sin \left(\theta \right)$
where $q$ is the charge of the particle, $v$ is its velocity, $B$ is the magnitude of the magnetic filed, and $\theta$ is the angle between the velocity and the magnetic field.

Given that the velocity is perpendicular to the magnetic field, we have that $\theta =0$, and consequently, $F=qvB$.

In this condition, the magnetic field offers a centripetal force on the particle.

Centripetal force is given as,
$F=\frac{m{v}^2}{r}$
where $m$ is the mass of the object, $v$ is its (tangential) velocity and $r$ is the radius of the circular path.

Equating the two equations of force, we get
$\begin{array}{rcl}& & \begin{array}{cc}& qvB=\frac{m{v}^2}{r}\\ \Rightarrow & r=\frac{mv}{qB}\end{array}\end{array}$

Step 2: Use relationship between momentum and kinetic energy

We know that Kinetic energy is given as
$K=\frac{1}{2}m{v}^2$
But, $p=mv$ where $p$ is the momentum of the particle. Thus,

$\begin{array}{cc}& p^2=m^2{v}^2\\ \Rightarrow & m{v}^2=\frac{p^2}{m}\\ \Rightarrow & K=\frac{1}{2}\frac{p^2}{m}\\ \Rightarrow & \frac{p^2}{m}=2K\\ \Rightarrow & p=\sqrt{2Km}\\ \Rightarrow & mv=\sqrt{2Km}\\ \Rightarrow & r=\frac{\sqrt{2Km}}{qB}\\ \Rightarrow & r\propto \frac{\sqrt{m}}{q}\end{array}$

Step 3: Determine mass of the particles

We know that protons and neutrons have identical masses with an error margin of $0.001$. If $m_p$ is the mass of a proton and $m_n$ is the mass of a neutron, we have
$m_p=m_n$
We know that a deuteron is the nucleus of a Deuterium atom (${}_1{}^2\mathrm{H}$ or $\mathrm{D}$), thus its mass is $2m_p$.
We know that the $\alpha$-particle is the nucleus of a ${}_2{}^4\mathrm{He}$ atom thus its mass is $4m_p$.

Step 4: Determine charge of the particles

As we know, the charge of a proton is $+1$.

The charge of a deuteron is also $+1$ (since it only has a proton and a neutron).

The charge of an $\alpha$-particle is $+2$ (since it has $2$ protons and $2$ neutrons).

Step 5: Calculate ratio of radii

The radius of the proton will be $r_p\propto \frac{\sqrt{m_p}}{1}$.

The radius of the deuteron will be $r_d\propto \frac{\sqrt{2m_p}}{1}$.

The radius of the $\alpha$-particle will be $r_{\alpha }\propto \frac{\sqrt{4m_p}}{2}$.

Thus, their ratio will be,

$\begin{array}{cc}& r_p:r_d:r_{\alpha }=\frac{\sqrt{m_p}}{1}:\frac{\sqrt{2m_p}}{1}:\frac{\sqrt{4m_p}}{2}\\ \therefore & r_p:r_d:r_{\alpha }=1:\sqrt{2}:2\end{array}$

Therefore, the the ratio of the radius of a proton, a deuteron, and an $\alpha$-particle moving perpendicular to a uniform magnetic field is $1:\sqrt{2}:2$.