Question

A proton of 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 the same radius?

• A. $1 \, MeV$
• B. $4 \, MeV$
• C. $2 \, MeV$
• D. $0.5 \, MeV$

Answer 1

Answer: (A)

$r=\frac{\sqrt{2 m K}}{q B}\Rightarrow K \propto \frac{q^{2}}{m}\Rightarrow \frac{K_{p}}{K_{\alpha }}=\left(\frac{q_{p}}{q_{\alpha }}\right)^{2}\times \frac{m_{\alpha }}{m_{p}}$
$\Rightarrow \frac{1}{K_{\alpha }}=\left(\frac{q_{p}}{2 q_{p}}\right)^{2}\times \frac{4 m_{p}}{m_{p}}=1\Rightarrow K_{\alpha }=1 \, MeV$ .