Question

A proton and an alpha particle having the same kinetic energy are allowed to pass through a uniform magnetic field perpendicular to the direction of their motion. Compare the radii of the paths of the proton and alpha particle.

Answer 1

Hint: Since both the proton and alpha particle is present in the magnetic field they experience Lorentz force. Also, the proton and alpha particle undertake a circular path, thus they experience centripetal force. We can use these two relations to get the radii of the proton and an alpha particle.

Formula used:
$KE_{p}=\dfrac{1}{2}mv^{2}$, $F=\dfrac{mv^{2}}{r}$ and $F=qvBsin\theta$

Complete answer:
Let the mass of the proton be $m$ and the charge be $q$. If the velocity of the proton is given by $v$, then the kinetic energy of the photon will be, $KE_{p}=\dfrac{1}{2}mv^{2}$.
Similarly. let the mass of alpha particle be $m\prime=4m$ and charge $q\prime=2q$.If the velocity of the alpha particle be $v\prime$, then the kinetic energy of the alpha particle will be, $KE_{a}=\dfrac{1}{2}m\prime v\prime ^{2}$

The Lorentz force is given as $F=qvBsin\theta$ where, $B$ is the magnetic field and $\theta$ is the angle between the direction of motion and the magnetic field.
The centripetal force is given by $F=\dfrac{mv^{2}}{r}$, where $r$ is the radius of the circular path.
For the proton and alpha particle to remain in the magnetic field, the Lorentz force is equal to the centripetal force . Then, $qvBsin\theta=\dfrac{mv^{2}}{r}$
$\Rightarrow qBsin\theta=\dfrac{mv}{r}$
$\Rightarrow r=\dfrac{mv}{qBsin\theta}$
Here, since the magnetic field perpendicular to the direction of their motion, or $\theta=90^{\circ}$.
$r=\dfrac{mv}{qBsin (90^{\circ})}=\dfrac{mv}{qB}$
Here, we can find the ratio between the radius of the photon $r_{p}$ and the radius of alpha particle $r_{a}$, using the above formula.
Substituting the values, we get $\dfrac{r_{p}}{r_{a}}=\dfrac{\dfrac{mv}{qB}}{\dfrac{m\prime v\prime}{q\prime B}}$.
Since the magnetic field $B$is constant, we get,$\dfrac{r_{p}}{r_{a}}=\dfrac{\dfrac{mv}{q}}{\dfrac{4m v\prime}{2q }}$

$\Rightarrow \dfrac{r_{p}}{r_{a}}=\dfrac{2v}{4 v\prime}$
$\Rightarrow \dfrac{r_{p}}{r_{a}}=\dfrac{v}{2 v\prime}$
Given that the kinetic energy of the photon is equal to the alpha particle.

Then,$KE_{p}=KE_{a}$
$\Rightarrow \dfrac{1}{2}mv^{2}=\dfrac{1}{2}m\prime v\prime ^{2}$
$\Rightarrow mv^{2}=4mv\prime^{2}$
$\Rightarrow \left(\dfrac{v}{v\prime}\right)^{2}=4$
$\Rightarrow \dfrac{v}{v\prime}=2$
We got a relation for $\dfrac{v}{v\prime}=2$, substituting them, we get, $\dfrac{r_{p}}{r_{a}}=\dfrac{2}{2}=1$
Hence, $r_{a}=r_{p}$

Note:
The mass of the alpha particle $m\prime$ is $4$ times the mass of the proton $m$ and the charge of the alpha particle $q\prime$ is 2 times the charge on the photon q. We are supposed to know these details. Due to the ratio in the charges and masses, they undertake the same radii.