A proton of energy 8 eV 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 (A) 4eV (B) 2eV (C) 8eV (D) 6eV

Given

Energy of a proton = 8eV

Here,

Momentum of a particle moving in a magnetic field mv = p = qBr

Hence, Kinetic energy of that particle can be written as KE = p2/2m = q2B2r2/2m

In the same magnetic field for the same path, KE ∝ q2/m

This ratio is equal for the alpha particle and the proton

(2e)2/4amu = 4e2/4amu = e2/4amu [Here, amu is the atomic mass unit]

Therefore, in such conditions both will have the same energy

So, energy of the alpha particle will be 8eV

Hence, the correct option is (C)

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