Question

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of $\alpha$ -particle (mass = 4 m 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​

Answer 1

The correct option is A 1 MeV​

By using $r=\frac{\sqrt{2mK}}{qB}$; $r\to$ same, $B\to$ same $\Rightarrow K\propto \frac{q^2}{m}$
Hence $\frac{K_{\alpha }}{K_p}={\left(\frac{q_{\alpha }}{q_p}\right)}^2\times \frac{m_p}{m_{\alpha }}={\left(\frac{2q_p}{q_p}\right)}^2\times \frac{m_p}{4m_p}=1\Rightarrow K_{\alpha }=K_p=1MeV$.