In a cyclotron, if a deutron can gain an energy of 40 MeV, then a proton can gain an energy of |
20 MeV 40 MeV 60 MeV 80 MeV |
80 MeV |
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: $80 \ \text{MeV}$ |