A proton and a deuteron execute circular orbits having same radii of 0.5 m in a plane perpendicular to a uniform magnetic field B. If the kinetic energy of the deuteron is 50 keV, the K.E. of the proton will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 100 keV
Radius of orbit: $r = \frac{mv}{qB}$
For same radius $r$ and same $q, B$:
$\frac{v_p}{v_d} = \frac{m_d}{m_p}$
Deuteron mass $m_d \approx 2m_p$
$\frac{v_p}{v_d} = 2$
Kinetic energy, $K = \frac{1}{2}mv^2$
$\frac{K_p}{K_d} = \frac{\frac{1}{2}m_p v_p^2}{\frac{1}{2}m_d v_d^2}$
$\frac{K_p}{K_d} = \frac{m_p (2v_d)^2}{2m_p v_d^2}$
$\frac{K_p}{K_d} = \frac{4}{2} = 2$
$K_p = 2K_d = 2 \times 50 \ \text{keV}$
$K_p = 100 \ \text{keV}$
Answer: 100 keV