R=qBmV=qB2m(kE)
Since R is same so KE∝mq2
So KE of α particle will be 4(2)2=1MeV
NEET UG 2015 — Physics Electromagnetism
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 circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:
Held on 30 Apr 2015 · Verified 9 Jul 2026.
0.5 MeV
1..5 MeV
1 MeV
4 MeV
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