A proton of mass m and charge +e is moving in a circular orbit in a magnetic field with energy 1 Me V. What should be the energy of α−particle (mass = 4m and charge = + 2e), so that it can revolve in the path of same radius?

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1 MeV

4 MeV

2 MeV

0.5 MeV

answer is A.

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Detailed Solution

r=2mKqB ⇒K ∝ q2m ⇒ KpKα = qpqα2 × mαmp ⇒ 1Kα = qp2qp2 × 4mpmp = 1 ⇒Kα =1 MeV

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