Question

A proton and an $\alpha -$ particle, having kinetic energies $K_p$ and $K_{\alpha }$ respectively, enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of $\alpha -$ particle is $2:1$. The ratio $K_p:K_{\alpha }$ will be :

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

Answer 1

The correct option is D $4:1$

Given :
$\frac{r_p}{r_{\alpha }}=\frac{2}{1}$

We know that,
$\frac{m_{\alpha }}{m_p}=\frac{4}{1}$ and $\frac{q_{\alpha }}{q_p}=\frac{2}{1}$

The radius of trajectory is given by,

$r=\frac{mv}{qB}=\frac{p}{qB}=\frac{\sqrt{2mK}}{qB}$

$\Rightarrow K=\frac{r^2{q}^2{B}^2}{2m}$

So,

$\frac{K_p}{K_{\alpha }}=\frac{m_{\alpha }{r}_p^2{q}_p^2}{m_p{r}_{\alpha }^2{q}_{\alpha }^2}=\frac{4}{1}\times {\left(\frac{2}{1}\right)}^2\times {\left(\frac{1}{2}\right)}^2=4:1$