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

Options

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

Answer

D

Explanation

Since , `K = (q^(2) r^(2) B^(2))/(2m) implies K prop (q^(2))/(m)`
`implies (K_(p))/(K_(d)) = ((q_(p))/(q_(d)))^(2) xx (m_(d))/(m_(p)) = ((1)/(2))^(2) xx ((2)/(1))`
`implies K_(p) = 2 xx 50 = 100`KeV