An electron, a proton and an alpha particle h | vedclass

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Question

As we know, radius of circular path in magnetic field $\mathrm{r}=\frac{\sqrt{2 \mathrm{Km}}}{\mathrm{qB}}$

For electron, $\mathrm{r}_{\mathrm{e}}=

Vedclass · Accepted Answer

$r_e < r_p$ $=$ $\;{r_\alpha }$

Vedclass · Answer

As we know, radius of circular path in magnetic field $\mathrm{r}=\frac{\sqrt{2 \mathrm{Km}}}{\mathrm{qB}}$

For electron, $\mathrm{r}_{\mathrm{e}}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{e}}}}{\mathrm{eB}}.........(i)$

For proton, $\mathrm{r}_{\mathrm{p}}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{p}}}}{\mathrm{eB}}.........(ii)$

For $\alpha$ particle,

$\mathrm{r}_{\alpha}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{a}}}}{\mathrm{q}_{\alpha} \mathrm{B}}=\frac{\sqrt{2 \mathrm{K} 4 \mathrm{m}_{\mathrm{p}}}}{2 \mathrm{eB}}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{p}}}}{\mathrm{eB}} .........(iii)$

$\mathrm{r}_{\mathrm{e}}<\mathrm{r}_{\mathrm{p}}=\mathrm{r}_{\alpha} \quad\left(\because \mathrm{m}_{\mathrm{e}}<\mathrm{m}_{\mathrm{p}}\right)$

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e,r_p$ and ${r_\alpha }$ respectively in a uniform magnetic field $B$. The relation between $r_e,r_p$ and $\;{r_\alpha }$ is

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