Question

A proton of energy $8eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

• A. $4eV$
• B. $2eV$
• C. $8eV$
• D. $6eV$

Answer 1

The correct option is C

$8eV$

Explanation for the correct option:

Step 1: Given

Energy of proton, $K_p=8eV$

Let $B$ be the magnetic field and $r$ be the radius of the circular path.
It is given that these are the same for both the proton and the $\alpha$-particle .

Step 2: Formulas used

We know that, the momentum of a particle moving in a magnetic field $B$ is, $p=mv=qBr$
where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $r$ is the radius of its circular path.

We also know that kinetic energy, $K=\frac{1}{2}m{v}^2$

Thus,

$\begin{array}{cc}& p^2=m^2{v}^2\\ \Rightarrow & m{v}^2=\frac{p^2}{m}\\ \Rightarrow & K=\frac{1}{2}\frac{p^2}{m}\\ \Rightarrow & K=\frac{q^2{B}^2{r}^2}{2m}\end{array}$

Step 3: Calculate the energy of $\alpha$-particle

The kinetic energy of the proton can be written as,
$K_p=\frac{{\left(q_p\right)}^2{B}^2{r}^2}{2m_p}$

An $\alpha$-particle is simply the nucleus of a helium-4 (${}_2{}^4\mathrm{He}$) atom.
Thus the mass of the $\alpha$-particle is $4$ times that of the proton i.e., $m_{\alpha }=4m_p$ ($m_{\alpha }$ is the mass of the $\alpha$-particle and $m_p$ is the mass of the proton)

The $\alpha$-particle has two protons thus its charge is twice that of a proton i.e., $q_{\alpha }=2q_p$ ($q_{\alpha }$ is the charge of the $\alpha$-particle and $q_p$ is that of the proton).

Thus,

$\begin{array}{cc}& K_{\alpha }=\frac{{\left(2q_p\right)}^2{B}^2{r}^2}{2\left(4m_p\right)}\\ \Rightarrow & K_{\alpha }=\frac{4{\left(q_p\right)}^2{B}^2{r}^2}{2\times 4m_p}\\ \Rightarrow & K_{\alpha }=\frac{{\left(q_p\right)}^2{B}^2{r}^2}{2m_p}\\ \Rightarrow & K_{\alpha }=K_p\\ \Rightarrow & K_{\alpha }=8eV\end{array}$

Hence, option C is correct.