Question

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

• A. 2 MeV
• B. 1 MeV
• C. 0.5 MeV
• D. 4 MeV

Answer 1

Answer: (B)

1 MeV

Answer 2

Kinetic energy of a charged particle,
$K=\frac{1}{2}mv^2$ or $v=\sqrt\frac{2K}{m}$
Radius of the circular path of a charged particle in
uniform magnetic field is given by
$R=\frac{mv}{Bq}=\frac{m}{Bq}\sqrt\frac{2K}{m}=\frac{\sqrt{2mK}}{Bq}$
Mass of a proton,$m_p=m$
Mass of an $\alpha$-particle, $\, m_{\alpha}=4m$
Charge of a proton, $q_p=e$
Charge of an $\alpha$-particle, $\, q_{\alpha}=2e$
$\therefore R_p=\frac{\sqrt{2m_p K_p}}{Bq_p}=\frac{\sqrt{2mK_p}}{Be}$
and $R_{\alpha}=\frac{\sqrt{2m_{\alpha} K_{\alpha}}}{Bq_{\alpha}}$
$=\frac{\sqrt{2(4m)K_{\alpha}}}{B(2e)}=\frac{\sqrt{2mK_{\alpha}}}{Be}$
$\therefore \frac{R_p}{R_{\alpha}}=\sqrt\frac{K_p}{K_{\alpha}}$
As $R_p=R_{\alpha}$ (given)
$\therefore K_{\alpha}=K_p=1 Me V$