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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......$MeV$
$1 $
$4$
$0.5$
$1.5$
Solution
The kinetic energy acquired by a charged particle in a uniform magnetic field $B$ is
$K=\frac{q^{2} B^{2} R^{2}}{2 m}\left(\text { as } R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}\right)$
where $q$ and $m$ are the charge and mass of the partic and $R$ is the radius of circular orbit
$\therefore $ The kinetic energy acquired by proton is
$K_{p}=\frac{q_{p}^{2} B^{2} R_{p}^{2}}{2 m_{p}}$
and that by the alpha particle is
$K_{\alpha}=\frac{q_{\alpha}^{2} B^{2} R_{\alpha}^{2}}{2 m_{\alpha}}$
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 $\quad K_{\alpha}=K_{p}\left(\frac{q_{\alpha}}{q_{p}}\right)^{2}\left(\frac{m_{p}}{m_{\alpha}}\right)\left(\frac{R_{\alpha}}{R_{p}}\right)^{2}$
Here, $K_{p}=1$ $MeV$ $, \frac{q_{\alpha}}{q_{p}}=2, \frac{m_{p}}{m_{\alpha}}=\frac{1}{4}$
and $\frac{R_{\alpha}}{R_{p}}=1$
$\therefore \quad K_{\alpha}=(1\, \mathrm{MeV})(2)^{2}\left(\frac{1}{4}\right)(1)^{2}=1 \,\mathrm{MeV}$
