Question
A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of α -particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius?
Answer: Option A
:
A
By using r=√2mKqB; r→ same, B→ same ⇒K∝q2m
Hence KαKp=(qαqp)2×mpmα=(2qpqp)2×mp4mp=1⇒Kα=Kp=1MeV.
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:
A
By using r=√2mKqB; r→ same, B→ same ⇒K∝q2m
Hence KαKp=(qαqp)2×mpmα=(2qpqp)2×mp4mp=1⇒Kα=Kp=1MeV.
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