Question

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

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

Answer 1

Answer: (A)

$r_{\alpha }=r_p<r_d$

We know,

$R=\frac{mv}{qB}=\frac{\sqrt{2mK.E}}{qB}$

When B,K.E=constant

$R\propto \frac{\sqrt{m}}{q}$

Let $m_p=m,q_p=q$

$r_{\alpha }\propto \frac{\sqrt{4m}}{2q}\propto \frac{2\sqrt{m}}{2q}\propto \frac{\sqrt{m}}{q}$

$r_d\propto \frac{\sqrt{2m}}{q}\propto \frac{\sqrt{2}\sqrt{m}}{q}$

$r_p\propto \frac{\sqrt{m}}{q}$

It is clear that,

$r_p=r_{\alpha }$

and , $r_d>r_p$

Hence option $\text{A}$ is the correct answer.