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 the trajectory of proton to that $\alpha -$particle is $2:1$. The ratio of $K_p:K_{\alpha }$ is :

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

Answer 1

The correct option is D

$4:1$

Step 1. Given Data,

A proton and an $\alpha -$particle, having kinetic energies $K_p$ and $K_{\alpha }$ respectively,

The ratio of the radii of the trajectory of proton to that $\alpha -$particle is $r_p:r_{\alpha }=$ $2:1$.

Step 2. Find the ratio of kinetic energy $K_p:K_{\alpha }$,

Since, We know that,

$$\begin{gathered} r=\frac{mv}{qB} \\ r=\frac{P}{qB} \end{gathered}$$

Where, $r=$radii of trajectory, $m=$mass, $v=$velocity of particle, $q=$ charge, $P=$ momentum and $B=$ magnetic flux density.

Let radii of proton to that $\alpha -$particle is $\frac{r_p}{r_{\alpha }}$ and kinetic energy be $\frac{K_p}{K_{\alpha }}$

Since we know that the charge ratio of proton and $\alpha -$particle is $\frac{q_{\alpha }}{q_p}=\frac{2}{1}$

Mass of alpha particle = 4 × mass of a proton

Mass ratio of proton and $\alpha -$particle is $\frac{m_{\alpha }}{m_p}=\frac{4}{1}$

Since,

$$\begin{gathered} \frac{r_p}{r_{\alpha }}=\frac{2}{1} \\ \frac{P_p}{P_{\alpha }}.\frac{q_{\alpha }}{q_p}=\frac{2}{1} \\ \frac{P_p}{P_{\alpha }}\times \frac{2}{1}=\frac{2}{1} \\ \frac{P_p}{P_{\alpha }}=1 \end{gathered}$$

Now, the ratio of kinetic energy is

$$\begin{gathered} \frac{K_p}{K_{\alpha }}=\left(\frac{P_p}{P_{\alpha }}\right).\left(\frac{m_{\alpha }}{m_p}\right) \\ \frac{K_p}{K_{\alpha }}=1.\frac{4}{1} \\ \frac{K_p}{K_{\alpha }}=\frac{4}{1} \end{gathered}$$

Therefore, $K_p:K_{\alpha }=4:1$

Hence, option D is correct.