A proton and an α-particle enter a uniform magnetic field B at right angles to the field. The radius of the circular orbit is the same for each. If the proton has kinetic energy 1 MeV, what is the kinetic energy of the α-particle?
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Question

A proton and an α-particle enter a uniform magnetic field B at right angles to the field. The radius of the circular orbit is the same for each. If the proton has kinetic energy 1 MeV, what is the kinetic energy of the α-particle?

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G1 Physics · Accepted Answer

Step 1: For a particle entering perpendicular to B, radius r=mv/(qB). Step 2: Kinetic energy K=(1/2)mv²=q²B²r²/(2m). Hence at equal B and r, K∝q²/m. Step 3: For proton (q=e,m=mₚ) and alpha (q=2e,m≈4mₚ), q²/m is identical, so K_α=Kₚ=1MeV; answer D. Final answer: (D) 1 MeV.