A proton, an electron, and a Helium nucleus, have the same energy.

Question:

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $\mathrm{r}_{\mathrm{p}}, \mathrm{r}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{He}}$ be their respective radii, then,

 

  1. (1) $\mathrm{r}_{\mathrm{e}}>\mathrm{r}_{\mathrm{p}}=\mathrm{r}_{\mathrm{He}}$

  2. (2) $\mathrm{r}_{\mathrm{e}}<\mathrm{r}_{\mathrm{p}}=\mathrm{r}_{\mathrm{He}}$

  3. (3) $\mathrm{r}_{\mathrm{e}}<\mathrm{r}_{\mathrm{p}}<\mathrm{r}_{\mathrm{He}}$

  4. (4) $\mathrm{r}_{\mathrm{e}}>\mathrm{r}_{\mathrm{p}}>\mathrm{r}_{\mathrm{He}}$


Correct Option: , 2

Solution:

(2) As $m v r=q v B \Rightarrow r=\frac{m v}{q B}=\frac{\sqrt{2 m K . E .}}{q B}$

$\left[\right.$ As $: \frac{1}{2} \mathrm{mv}^{2}=\mathrm{K} . \mathrm{E} .$

$\Rightarrow \mathrm{m}^{2} \mathrm{v}^{2}=2 \mathrm{~m} \mathrm{~K} . \mathrm{E} .$

$\Rightarrow \mathrm{mv}=\sqrt{2 \mathrm{~m} \mathrm{~K} . \mathrm{E} .}]$

For proton, electron and $\alpha$-particle,

$\mathrm{m}_{\mathrm{He}}=4 \mathrm{~m}_{\mathrm{p}}$ and $\mathrm{m}_{\mathrm{p}} \gg \mathrm{m}_{\mathrm{e}}$

Also $\mathrm{a}_{\mathrm{He}}=2 \mathrm{q}_{\mathrm{p}}$ and $\mathrm{q}_{\mathrm{p}}=\mathrm{q}_{\mathrm{e}}$

$\therefore$ As $\mathrm{KE}$ of all the particles is same then,

$\mathrm{r} \alpha \frac{\sqrt{\mathrm{m}}}{\mathrm{q}}$

$\therefore \mathrm{r}_{\mathrm{He}}=\mathrm{r}_{\mathrm{p}}>\mathrm{r}_{\mathrm{e}}$

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