A proton and an alpha – partical,
Question:

A proton and an $\alpha$-particle, having kinetic energies $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\alpha}$, 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 of $\mathrm{K}_{\mathrm{p}}: \mathrm{K}_{\alpha}$ is:

1. $1: 8$

2. $8: 1$

3. $1: 4$

4. $4: 1$

Correct Option: , 4

Solution:

$r=\frac{m v}{q B}=\frac{p}{q B} \quad \frac{m_{\alpha}}{m_{p}}=4$

$\frac{r_{p}}{r_{\alpha}}=\frac{p_{p}}{q_{p}} \frac{q_{\alpha}}{p_{\alpha}}=\frac{2}{1}$

$\frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\alpha}}=\frac{2 \mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\alpha}}=2\left(\frac{1}{2}\right)$

$\frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\alpha}}=1$

$\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\alpha}}=\frac{\mathrm{p}_{\mathrm{p}}^{2}}{\mathrm{p}_{\alpha}^{\mathrm{p}}} \frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{p}}}=$