Question

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

Answer 1

Correct Answer - 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`.