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Question

A deutron of kinetic energy 50keV is describing a circular orbit of radius 0.5m in a plane perpendicular to magnetic field B . The kinetic energy of the proton that describes a circular orbit of radius 0.5m in the same plane with the same B is :

A
25keV
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B
50keV
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C
200keV
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D
100keV
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Solution

The correct option is C 100keV
Deutron is 1H2
r=mvqB
r=2K.mqB
r1=r2
2×50×2B=2×K.1B
K=100keV

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