JEE Advanced 2022 Paper 1 · Q02 · Nuclear Reactions
The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ⁷N¹⁶ + ₂He⁴ → ₁H¹ + ₈O¹⁹ in a laboratory frame is n (in MeV). Assume that ⁷N¹⁶ is at rest in the laboratory frame. The masses of ⁷N¹⁶, ₂He⁴, ₁H¹ and ₈O¹⁹ can be taken to be 16.006 u, 4.003 u, 1.008 u and 19.003 u, respectively, where 1 u = 930 MeV·c⁻². The value of n is ___.
Reveal answer + step-by-step solution
Correct answer:2.32
Solution
Q-value of the reaction: Q = (m_N + m_He − m_H − m_O)·c² = (16.006 + 4.003 − 1.008 − 19.003)·930 MeV = (−0.002)·930 MeV = −1.86 MeV (endothermic). In the lab frame with target ⁷N¹⁶ at rest, the threshold kinetic energy of the incident alpha is K_th = |Q|·(m_α + m_N)/m_N = 1.86·(4.003 + 16.006)/16.006 = 1.86·20.009/16.006 ≈ 1.86·1.250 ≈ 2.325 MeV. Rounded to two decimals: n = 2.32.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →