Question

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

We know that the radius of circular track in a magnatic field $=\frac{mv}{qB}$.......(i)
Momentum in terms of kinetic field is given by
$P=\sqrt{2mK}$ = $mv$ .......(ii)
Using (ii) in (i) we get
$r=\frac{\sqrt{2mK}}{qB}$

As all these particles have same kinetic energy and are under same magnetic field so we can say
$r\propto \frac{\sqrt{m}}{q}$

By using this we can write
$r_{\alpha }:r_d:r_P=\frac{\sqrt{4m}}{2q}:\frac{\sqrt{2m}}{q}:\frac{\sqrt{m}}{q}$
$⟹r_{\alpha }:r_d:r_P=1:\sqrt{2}:1$

On comparing this we get
$r_{\alpha }=r_P<r_d$