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 $1MeV$, the energy acquired by the alpha particle will be

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

Answer 1

Answer: (B)

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 $R=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}$)
where $q$ and m are the charge and mass of the particle and $R$ is the radius of circular orbit.
$\therefore$ 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,$\frac{K_{\alpha }}{K_p}={\left(\frac{q_{\alpha }}{q_p}\right)}^2\left(\frac{m_p}{m_{\alpha }}\right){\left(\frac{R_{\alpha }}{R_p}\right)}^2$
or $K_{\alpha }=K_p{\left(\frac{q_{\alpha }}{q_p}\right)}^2\left(\frac{m_p}{m_{\alpha }}\right){\left(\frac{R_{\alpha }}{R_p}\right)}^2$
Here, $K_p=1MeV,\frac{q_{\alpha }}{q_p}=2,\frac{m_p}{m_{\alpha }}=\frac{1}{4}$
and $\frac{R_{\alpha }}{R_p}=1$
$\therefore K_{\alpha }=(1MeV)(2{)}^2\left(\frac{1}{4}\right)(1{)}^2=1MeV$