Question

A proton and an $\alpha$ - particle (Helium nucleus) enters a uniform magnetic field with their direction of velocity perpendicular to the direction of magnetic field.

$\left(iv\right)$ If proton and $\alpha$ - particle enter the magnetic field, perpendicular to it with same kinetic energy, then the ratio of the radii of curvature of proton and $\alpha$ - particle will be

​​​​​​​$[$1 Mark$]$

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

Answer 1

The correct option is D $1:1$

When $v$ and $B$ are perpendicular, the path will be circle.

The radius of the circle is given by $r=\frac{mv}{qB}$

For proton, let $M_P=M,q_P=q$
Radius of proton is
$r_P=\frac{\sqrt{2m\left(KE\right)}}{qB}$

For $\alpha$ - particle, $M_{\alpha }=4m,q_{\alpha }=2q$
Radis of alpha particle is
$r_{\alpha }=\frac{\sqrt{2\left(4m\right)\left(KE\right)}}{2qB}$

$\Rightarrow r_{\alpha }=\frac{2\sqrt{2m\left(KE\right)}}{2qB}$

$\Rightarrow r_{\alpha }=\frac{\sqrt{2m\left(KE\right)}}{qB}$

Ratio of radii
$\therefore \frac{r_P}{r_{\alpha }}=\frac{1}{1}$