Question

A proton and an alpha particle both enter a region of uniform magnetic field $B$, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 $MeV$, the energy acquired by the alpha particle will be

• A. 1.5 $MeV$
• B. 1 $MeV$
• C. 4 $MeV$
• D. 0.5 $MeV$

Answer 1

The correct option is B 1 $MeV$

The kinetic energy acquired by a charged particle in a uniform magnetic field $B$ is
$K=\frac{q^2{B}^2{R}^2}{2m}$ (as $R=\frac{mv}{qB}=\sqrt{\frac{2mK}{qB}}$)
where $q$ and $m$ are the charge and mass of the particle and $R$ is the radius of circular orbit.
∴ ​​​​​​The kinetic energy acquired by proton is
$K_p=\frac{q_p^2{B}^2{R}_p^2}{2m_p}$
and that by the alpha particle is

$K_{\alpha }=\frac{q_{\alpha }^2{B}^2{R}_{\alpha }^2}{2m_{\alpha }}$
Thus,
$\frac{K_{\alpha }}{K_p}=\frac{q_{\alpha }^2{R}_{\alpha }^2{m}_p}{q_p^2{R}_p^2{m}_{\alpha }}$
Here, $K_p$=$1MeV$, $\frac{q_{\alpha }}{q_p}=2,\frac{m_p}{m_{\alpha }}=\frac{1}{4},\frac{R_{\alpha }}{R_p}=1$

∴ $K_{\alpha }=\left(1MeV\right)\left(2{)}^2\right(\frac{1}{4}\left)\right(1{)}^2=1MeV$.