Question

A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m, in a plane perpendicular to magnetic field $\overrightarrow{B}$. The kinetic energy of a proton that discribes a circular orbit of radius 0.5 m in the same plane with the same magnetic field $\overrightarrow{B}$ is

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

Answer 1

The correct option is C 100 keV

$r_{deutron}=\frac{\sqrt{2m_d{E}_d}}{Bq};r_{proton}=\frac{\sqrt{2m_p{E}_p}}{Bq}$

For same radius, B and q:

$m_p{E}_p=m_d{E}_d$

$\Rightarrow E_p=\frac{m_d}{m_p}{E}_d=\frac{2}{1}\times 50=100keV$