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

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4 eV

2 eV

8 eV

6 eV

answer is C.

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Detailed Solution

r=2mKqB ⇒q ∝ mK ⇒ K ∝ q2m⇒KαKp = qαqp2 × mpmα ⇒ Kα8 = 2qpqp2 × mp4mp=1⇒ Kα = 8 eV

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