Question

An $\alpha$-particle and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is:

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $1:8$

Answer 1

The correct option is A $1:1$

Kinetic energy of $\alpha$ particle is equal to the kinetic energy of proton.

$\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2=\frac{1}{2}{m}_p{v}_p^2$

We know $m_{\alpha }=4m_p$

$⟹v_p=2v_{\alpha }$

Magnetic force acting on a particle in a magnetic field provides the necessary centripetal force for circular motion.

$\frac{m{v}^2}{R}=qvB$

$R=\frac{mv}{qB}$

Therefore, ratio of radii of the paths of $\alpha$ particle to that of proton is:

$\frac{R_{\alpha }}{R_p}=\frac{\frac{m_{\alpha }{v}_{\alpha }}{q_{\alpha }B}}{\frac{m_p{v}_p}{q_{p}B}}$

$m_{\alpha }=4m_p;v_p=2v_{\alpha };q_{\alpha }=2q_p$

Substituting values,

$R_{\alpha }:R_p=1:1$