Question

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

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

Answer 1

Answer: (D)

$100keV$

Deutron is ${}_1{H}^2$
$r=\frac{mv}{qB}$
$r=\frac{\sqrt{2K.m}}{qB}$
$\because r_1=r_2$
$\frac{\sqrt{2\times 50\times 2}}{B}=\frac{\sqrt{2\times K.1}}{B}$
$K=100keV$