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Q.

In a nuclear experiment a 1 Mev proton moves in a uniform magnetic field in a circular path of a certain radius.The energy of a deuteron which will circulate in the same orbit in the same magnetic field will be

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a

1 Mev

b

4 Mev

c

0.5 Mev

d

2 Mev

answer is C.

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Detailed Solution

ω=QBm,Speed  v=2km∴K= 12mω2R2=Q2B2R22m∴ kdkp=Qd2Qp2×mpmd=(1)2×12=12∴ kd=0.5 Mev
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In a nuclear experiment a 1 Mev proton moves in a uniform magnetic field in a circular path of a certain radius.The energy of a deuteron which will circulate in the same orbit in the same magnetic field will be