A proton of energy 8eV is moving in a circular path in a uniform magnetic field. The energy of an α- particle moving in the same magnetic field and along the same path will be-

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4eV

2eV

8eV

6eV

answer is C.

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Detailed Solution

r=mvqB=pqB=2mkqB p22m=k; p=2mkIt r and B are samek∝qm kpkα=q2q4mm=1 kp=kα.

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