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Q.

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

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a

25 keV

b

50 keV

c

200 keV

d

100 keV

answer is D.

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Detailed Solution

r=2mKqB  ⇒K ∝  q2m⇒KpKd  = qpqd2 × mdmp = 112  ×  21=21⇒ Kp=2  ×  50  = 100  keV
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