Question

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 $\vec { B }$ is

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

Answer 1

Answer: (D)

100 keV

Answer 2

$r = \frac { \sqrt { 2 m K } } { q B } \Rightarrow K \propto \frac { q ^ { 2 } } { m }$

$\Rightarrow \frac { K _ { p } } { K _ { d } } = \left( \frac { q _ { p } } { q _ { d } } \right) ^ { 2 } \times \frac { m _ { d } } { m _ { p } } = \left( \frac { 1 } { 1 } \right) ^ { 2 } \times \frac { 2 } { 1 } = \frac { 2 } { 1 }$

$\Rightarrow K _ { p } = 2 \times 50 = 100 \mathrm { keV }$