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 maximum energy gained by a charged particle is
$E = \frac{q^2 B^2 R^2}{2m}$
For two particles in the same cyclotron (same $q, B, R$):
$E \propto \frac{1}{m}$
Given: $E_d = 40 \ \text{MeV}$ for deuteron, with $m_d \approx 2m_p$.
So, $E_p = \frac{m_d}{m_p} \cdot E_d = \frac{2m_p}{m_p} \cdot 40 = 80 \ \text{MeV}$
Answer: $80 \ \text{MeV}$