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← JEE Advanced 2022 · Paper 1
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JEE Advanced 2022 · Paper 1, Physics question 2

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction 716N+24He→11H+819O{}^{16}_{7}\mathrm N + {}^{4}_{2}\mathrm{He} \to {}^{1}_{1}\mathrm H + {}^{19}_{8}\mathrm O in a laboratory frame is nn (in MeV). Assume that 716N{}^{16}_{7}\mathrm N is at rest in the laboratory frame. The masses of 716N{}^{16}_{7}\mathrm N, 24He{}^{4}_{2}\mathrm{He}, 11H{}^{1}_{1}\mathrm H and 819O{}^{19}_{8}\mathrm O can be taken to be 16.006 uu, 4.003 uu, 1.008 uu and 19.003 uu, respectively, where 1 u=930 MeV c−21\,u = 930\ \mathrm{MeV}\,c^{-2}. The value of nn is _____.

Answer and worked solution

Official answer: 2.3

Q-value.
Δm=(16.006+4.003)−(1.008+19.003)=−0.002 u,\Delta m=(16.006+4.003)-(1.008+19.003)=-0.002\,u,
so Q=−0.002×930=−1.86Q=-0.002\times930=-1.86 MeV (endothermic).

Threshold. Momentum conservation leaves part of the alpha's energy as kinetic energy of the products. The minimum alpha energy is
Kmin⁡=∣Q∣(1+mαmN)=1.86(1+4.00316.006)=2.33 MeV.K_{\min}=|Q|\left(1+\frac{m_\alpha}{m_N}\right)=1.86\left(1+\frac{4.003}{16.006}\right)=2.33\ \text{MeV}.