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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?

A
1 MeV​
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B
4 MeV​
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C
2 MeV​
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D
0.5 MeV​
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Solution

The correct option is A 1 MeV​
By using r=2mKqB; r same, B same Kq2m
Hence KαKp=(qαqp)2×mpmα=(2qpqp)2×mp4mp=1Kα=Kp=1MeV.

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