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Q.

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1 MeV. What should be the energy of α-particle (mass 4 m and charge + 2 e) so that it can revolve in the path of same radius

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a

1 MeV

b

4 MeV

c

2 MeV

d

0 .5 MeV

answer is A.

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Detailed Solution

we know that r=mvqB  or  r2=m2v2q2B2In terms of energy, the above equation can be written asr2=2m×1/2mv2q2B2=2mEq2B2Here r and B are same, hence    m1E1q12=m2E2q22 or m×1e2=3m×E24e2∴     E2=1MeV
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