Question

A proton, a deutron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant uniform magnetic field. If $r_p,r_d,r_{\alpha }$ denote respectively the radii of trajectories of these particles, then -

• A. $r_{\alpha }=r_p<r_d$
• B. $r_{\alpha }>r_d>r_p$
• C. $r_{\alpha }=r_d>r_p$
• D. $r_p=r_d=r_{\alpha }$

Answer 1

The correct option is A $r_{\alpha }=r_p<r_d$

Given that $K_p=K_d=K_a=K$

We know that $m_p=m,m_d=2m$

and $m_{\alpha }=4m$

and $q,=e,q=e$

and $q_{\alpha }=2e$

Further $r=\frac{\sqrt{2mK}}{qB}$

$\Rightarrow r_p=\frac{\sqrt{2mK}}{eB}$

$r_d=\frac{\sqrt{2\left(2m\right)K}}{eB}=\sqrt{2}{r}_p$

and $r_{\alpha }=\frac{\sqrt{2\left(4m\right)K}}{\left(2e\right)B}=r_p$.

Hence $r_{\alpha }=r_p<r_d$