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

Answer: (B)

The kinetic energy acquired by a charged
particle in a uniform magnetic field $B$ is
$K=\frac{q^2B^2R^2}{2m}$ (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
$K_p=\frac{q_p^2B^2R_p^2}{2m_p}$
and that by the alpha particle is
$K_{\alpha}=\frac{q_{\alpha}^2B^2R_{\alpha}^2}{2m_{\alpha}}$
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, $K_p=1 \, MeV, \frac{q_{\alpha}}{q_p}=2,\frac{m_p}{m_{\alpha}}=\frac{1}{4}$
and $\frac{R_{\alpha}}{R_p}=1$
$\therefore K_{\alpha}=(1 \, MeV)(2)^2 \left(\frac{1}{4}\right)(1)^2=1 \, MeV$