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

50 keV

$ R = \frac{mv}{qB} = \frac{\sqrt {2mE}}{qB}$

$\frac{R_p}{R_d} = \sqrt {\frac{m_p}{m_d}} \sqrt {\frac{E_p}{E_d}} $

$ 1 = \sqrt{\frac{1}{2}} \sqrt{\frac{E_p}{50KeV}}$

$ E_p = 2\times 50KeV = 100KeV$