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)

$1\, MeV$

Answer 2

The kinetic energy acquired by a charged
particle in a uniform magnetic field B is
\, \, \, K=\frac{q^2B^2R^2}{2m} \hspace20mm \big(as \, R=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}\big)
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,$\frac{K_{\alpha}}{K_p}=\big(\frac{q_{\alpha}}{q_p}\big)^2\big(\frac{m_p}{m_{\alpha}}\big) \big(\frac{R_{\alpha}}{R_p}\big)^2$
or $K_{\alpha}=K_p\big(\frac{q_{\alpha}}{q_p}\big)^2\big(\frac{m_p}{m_{\alpha}}\big) \big(\frac{R_{\alpha}}{R_p}\big)^2$
Here, $K_p=1 \, MeV, \frac{q_{\alpha}}{q_p}=2,\frac{m_p}{m_{\alpha}}=\frac{1}{4}$
and $\frac{R_{\alpha}}{R_p}=1$
$\therefore \, \, K_{\alpha}=(1 \, MeV)(2)^2 \big(\frac{1}{4}\big)(1)^2=1 \, MeV$