A proton and an alpha-particle,having kinetic | vedclass

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(D) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(D) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum.Given the ratio of radii $\frac{r_{p}}{r_{\alpha}} = \frac{2}{1}$.We know that for an $\alpha$-particle,mass $m_{\alpha} = 4m_{p}$ and charge $q_{\alpha} = 2q_{p}$.Using the formula $\frac{r_{p}}{r_{\alpha}} = \frac{p_{p}}{q_{p}B} \cdot \frac{q_{\alpha}B}{p_{\alpha}} = \frac{p_{p}}{p_{\alpha}} \cdot \frac{q_{\alpha}}{q_{p}} = 2$.Substituting the charge ratio: $\frac{p_{p}}{p_{\alpha}} \cdot \frac{2q_{p}}{q_{p}} = 2 \implies \frac{p_{p}}{p_{\alpha}} \cdot 2 = 2 \implies \frac{p_{p}}{p_{\alpha}} = 1$.Kinetic energy $K$ is related to momentum $p$ by $K = \frac{p^2}{2m}$.Therefore,$\frac{K_{p}}{K_{\alpha}} = \frac{p_{p}^2}{2m_{p}} \cdot \frac{2m_{\alpha}}{p_{\alpha}^2} = \left(\frac{p_{p}}{p_{\alpha}}\right)^2 \cdot \frac{m_{\alpha}}{m_{p}}$.Substituting the values: $\frac{K_{p}}{K_{\alpha}} = (1)^2 \cdot \frac{4m_{p}}{m_{p}} = 1 \cdot 4 = 4$.Thus,the ratio $K_{p}:K_{\alpha}$ is $4:1$.

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

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