Question

A proton, a deuteron and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths described by them is

• A. $\sqrt{2}:1:\sqrt{2}$
• B. $1:\sqrt{2}:1$
• C. $\sqrt{2}:\sqrt{2}:1$
• D. $\sqrt{2}:1:1$

Answer 1

The correct option is A $\sqrt{2}:1:\sqrt{2}$

Radius of circular path $r=\frac{mv}{Bq}$

Using $p=mv=\sqrt{2mK}$

where $K$ is the kinetic energy of the particle.

We get $r=\frac{\sqrt{2mK}}{Bq}$

$⟹$ $r\propto \frac{\sqrt{m}}{q}$

Given : $q_{\alpha }=2q_p$ and $q_d=2q_p$ where $q_p$ is the charge of proton.

Also, $m_{\alpha }=4m_p$ and $m_d=2m_p$ where $m_p$ is the mass of proton.

$\therefore$ $r_p:r_d:r_{\alpha }=\frac{\sqrt{m_p}}{q_p}:\frac{\sqrt{m_d}}{q_d}:\frac{\sqrt{m_{\alpha }}}{q_{\alpha }}$

Or $r_p:r_d:r_{\alpha }=\frac{\sqrt{m_p}}{q_p}:\frac{\sqrt{2m_p}}{2q_p}:\frac{\sqrt{4m_p}}{2q_p}$

$⟹$ $r_p:r_d:r_{\alpha }=\sqrt{2}:1:\sqrt{2}$