In a cyclotron, if a deutron can gain an energy of 40 MeV, then a proton can gain an energy of
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 80 MeV
In a cyclotron the particle leaves at the radius \( R \) of the dees, where \( qvB = mv^{2}/R \), so \( v = qBR/m \). The maximum kinetic energy is
\( \displaystyle K = \frac{1}{2}mv^{2} = \frac{q^{2}B^{2}R^{2}}{2m} \)
For the same cyclotron \( B \) and \( R \) are fixed. A deutron and a proton carry the same charge \( e \), so \( K \propto 1/m \).
\( \displaystyle \frac{K_{p}}{K_{d}} = \frac{m_{d}}{m_{p}} = 2 \)
\( \displaystyle K_{p} = 2\times40\,\mathrm{MeV} = 80\,\mathrm{MeV} \)
The lighter proton gains more energy.
Note: The same question was also asked in CUET (UG) 2025 on 13-05-2025 (Slot 1).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Cyclotron – “… From Eq. (4.5) we have,
\( \displaystyle v = \frac{qBR}{m} \qquad (4.9) \)
where R is the radius of the trajectory at exit, and equals the radius of a dee.
Hence, the kinetic energy of the ions is,
\( \displaystyle \frac{1}{2}mv^{2} = \frac{q^{2}B^{2}R^{2}}{2m} \qquad (4.10) \)”