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Q.

An α-particle and a proton are projected in a uniform magnetic field in a direction perpendicular to the field. If r be the radius of circular path energy of α-particle to that of proton is

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a

1 : 4

b

2 : 1

c

4 : 1

d

1 : 2

answer is C.

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Detailed Solution

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kinetic energy, $K = \frac{1}{2} {mv}^{2} \Rightarrow v = \sqrt{\frac{2 K}{m}}$

$∴ r = \frac{v}{ω} = \frac{\sqrt{2 K / m}}{QB / m} = \frac{\sqrt{2 Km}}{QB} \Rightarrow K = \frac{Q^{2} B^{2} r^{2}}{2 m} \Rightarrow K \propto \frac{Q^{2} r^{2}}{m} ∴ \frac{K_{α}}{K_{p}} = {(\frac{Q_{α}}{Q_{p}})}^{2} . {(\frac{r_{α}}{r_{p}})}^{2} . {(\frac{m_{p}}{m_{α}})}^{2} = {(\frac{2 e}{e})}^{2} . {(2)}^{2} . (\frac{m}{4 m}) = 4$

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