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A proton and an alpha particle both enter a region of uniform magnetic field, B, moving at
right angles to the field B. If the radius of circular orbits for both the particles is equal and the
kinetic energy acquired by proton is 1MeV, the energy acquired by the alpha particle will be

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1.5 MeV

4 MeV

0.5 MeV

1 MeV

answer is A.

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Detailed Solution

$R = \frac{mv}{q_{B}} = \frac{\sqrt{2mK}}{q_{B}} ∵ R_{α} = R_{p} ∴ \frac{4 m_{α} k_{a}}{q_{α}^{2} B^{2}} = \frac{{4m}_{p} K_{p}}{q_{p}^{2} B^{2}} = \frac{4 m_{p} k_{α}}{4 e^{2}} = \frac{m_{p} (1MeV)}{e^{2}} \Rightarrow K_{α} = 1 MeV$

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