A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of the circular orbits for both particles is equal and the kinetic energy acquired by the proton is 1 MeV, the energy acquired by the alpha particle will be

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(2) 1 MeV. Given: equal radii in the same field; the alpha particle has twice the charge and four times the mass of the proton. Idea: write the kinetic energy in terms of the radius, which is what is held fixed. From r=(mv)/(qB): mv=qBr, so v=(qBr)/m. K=1/2mv²=(q²B²r²)/(2m). With B and r the same for both, K∝(q²)/m. (K_α)/(Kₚ)=(q_α²)/(qₚ²)×(mₚ)/(m_α)=4×1/4=1. So the alpha particle also acquires 1 MeV.

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of the circular orbits for both particles is equal and the kinetic energy acquired by the proton is 1 MeV, the energy acquired by the alpha particle will be

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