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 of the proton is $1MeV$, the energy of the alpha particle will be

• A. $1MeV$
• B. $4MeV$
• C. $0.5MeV$
• D. $1.5MeV$

Answer 1

The correct option is A $1MeV$

Radius of circular orbit of the particle is given by
$R=\frac{mv}{qB}=\frac{\sqrt{2mE}}{qB}$
Since $R$ is same for proton and alpha particle
$\therefore E\propto \frac{q^2}{m}$
$\frac{E_{\alpha }}{E_p}=\frac{q_{\alpha }^2}{m_{\alpha }}\times \frac{m_p}{q_p^2}=\frac{(2q_p{)}^2}{4m_p}\times \frac{m_p}{q_p^2}$
$\Rightarrow E_{\alpha }=E_p=1MeV$