Q. A proton a mass $m$ and charge $+ e$ is moving in a circular orbit in a magnetic field with energy $1\ MeV$. What should be the energy of $\alpha -$particle (mass $= 4m$ and charge $= + 2e$), so that it can revolve in the path of same radius
A) $1\ MeV$
B) 4 $MeV$
C) 2 $MeV$
D) 0.5 $MeV$
Explanation
(a) $r = \frac{\sqrt{2mK}}{qB} \Rightarrow K \propto \frac{q^{2}}{m} \Rightarrow \frac{K_{p}}{K_{\alpha}} = \left( \frac{q_{p}}{q_{\alpha}} \right)^{2} \times \frac{m_{\alpha}}{m_{p}}$
$\Rightarrow \frac{1}{K_{\alpha}} = \left( \frac{q_{p}}{2q_{p}} \right) \times \frac{4m_{p}}{m_{p}} = 1 \Rightarrow K_{\alpha} = 1\ MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.