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
The magnetic force \( qvB \) provides the centripetal force, so the radius of the circular orbit is
\( \displaystyle r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} \)
The proton and the deuteron have the same charge \( e \), move in the same field \( B \) and have the same radius, so \( \sqrt{2mK} \) is the same for both:
\( m_{p} K_{p} = m_{d} K_{d} \)
The deuteron has twice the mass of the proton, \( m_{d} = 2m_{p} \):
\( \displaystyle K_{p} = K_{d}\,\frac{m_{d}}{m_{p}} = 50\,\mathrm{keV} \times 2 = 100\,\mathrm{keV} \)