An -particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. W

Question

An $α$ -particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $α$ -particle and the sulfur ion move in circular orbits of radii $r_{α}$ and $r_{s}$ , respectively. The ratio $(\frac{r_{s}}{r_{α}})$ is __________. [2021]

Smart Achievers · Accepted Answer

(4)

$Here, force F = q V B is balanced by centripetal force$

$F_{e} = \frac{m V^{2}}{r}$

$∴$ $q V B = \frac{m V^{2}}{r}$

or $r = \frac{m V}{q B}$ $= \frac{\sqrt{2 m q V}}{q B}$

$\frac{p^{2}}{2 m} = K.E. = q V$

$p = \sqrt{2 m q V}$

Therefore, $r = \frac{p}{q B} = \frac{\sqrt{2 m q V}}{q B}$

$\frac{r_{s}}{r_{α}} = \sqrt{\frac{m_{s}}{q_{s}}} \times \sqrt{\frac{q_{α}}{m_{α}}}$ $= \sqrt{\frac{32}{1}} \times \sqrt{\frac{2}{4}} = 4$

$∴$ $\frac{r_{s}}{r_{α}} = 4$

Solution

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Solution

Ans. (4) Here, force F=qVB is balanced by centripetal force Fe=mV2r ∴ qVB=mV2r or r=mVqB=2mqVqB p22m=K.E.=qV p=2mqV Therefore, r=pqB=2mqVqB rsrα=msqs×qαmα=321×24=4 ∴ rsrα=4

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