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Nuclear Reactions

Problem 6.258

Find the energy required for separation of a \(\mathrm{Ne}^{20}\) nucleus into two alpha-particles and a \(\mathrm{C}^{12}\) nucleus if it is known that the binding energies per one nucleon in \(\mathrm{Ne}^{20}, \mathrm{He}^{4},\) and \(\mathrm{C}^{12}\) nuclei are equal to \(8.03,7.07,\) and \(7.68 \mathrm{MeV}\) respectively.

Reveal Answer
E=20εNe−2⋅4εα−12εC=11.9MeV, where ε is the binding energy per nucleon in the corresponding nucleusE=20 \varepsilon_{\mathrm{Ne}}-2 \cdot 4 \varepsilon_{\alpha}-12 \varepsilon_{\mathrm{C}}=11.9 \mathrm{MeV}, \text { where } \varepsilon \text { is the binding energy per nucleon in the corresponding nucleus}
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