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

$4eV$

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B

$2eV$

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C

$8eV$

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D

$6eV$

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Solution

## The correct option is C $8eV$Explanation for the correct option:Step 1: GivenEnergy 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 usedWe 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,$\begin{array}{cc}& {p}^{2}={m}^{2}{v}^{2}\\ ⇒& m{v}^{2}=\frac{{p}^{2}}{m}\\ ⇒& K=\frac{1}{2}\frac{{p}^{2}}{m}\\ ⇒& K=\frac{{q}^{2}{B}^{2}{r}^{2}}{2m}\end{array}$Step 3: Calculate the energy of $\alpha$-particleThe kinetic energy of the proton can be written as,${K}_{p}=\frac{{\left({q}_{p}\right)}^{2}{B}^{2}{r}^{2}}{2{m}_{p}}$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., ${m}_{\alpha }=4{m}_{p}$ (${m}_{\alpha }$ 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., ${q}_{\alpha }=2{q}_{p}$ (${q}_{\alpha }$ is the charge of the $\alpha$-particle and ${q}_{p}$ is that of the proton).Thus,$\begin{array}{cc}& {K}_{\alpha }=\frac{{\left(2{q}_{p}\right)}^{2}{B}^{2}{r}^{2}}{2\left(4{m}_{p}\right)}\\ ⇒& {K}_{\alpha }=\frac{4{\left({q}_{p}\right)}^{2}{B}^{2}{r}^{2}}{2×4{m}_{p}}\\ ⇒& {K}_{\alpha }=\frac{{\left({q}_{p}\right)}^{2}{B}^{2}{r}^{2}}{2{m}_{p}}\\ ⇒& {K}_{\alpha }={K}_{p}\\ ⇒& {K}_{\alpha }=8eV\end{array}$Hence, option C is correct.

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