A deutron of kinetic energy `50` keV is describing a circular orbit of radius `0.5` meter in a plane perpendicular to magnetic field `vecB`. The kinetic energy of the proton that describes a circular orbit of radius `0.5` meter in the same plane with the same `vecB` is
A
200 keV
B
50 keV
C
100keV
D
25 keV
Text Solution
Verified by Experts
The correct Answer is:
C
`E=(q^2B^2r^2)/(2m)` `therefore (E_P)/(E_d)=((q_d)/(q_p))^2(m_P)/(m_d)=2` `therefore E_P= 2 E_d =2xx50=100 k e V`.
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