Physics · 2019
JEE Main · 10 January 2019, Shift 2 · Q26
Consider the nuclear fission Ne^20 → 2 He^4+ C^12 Given that the binding energy/nucleon of Ne^20, He^4 and C^12 are, respectively, 8.03 MeV, 7.07 MeV…
Consider the nuclear fission
$$\mathrm{Ne}^{20} \rightarrow 2 \mathrm{He}^4+\mathrm{C}^{12}
$$
Given that the binding energy/nucleon of $\displaystyle \mathrm{Ne}^{20}, \mathrm{He}^4$ and $\displaystyle \mathrm{C}^{12}$ are, respectively, $\displaystyle 8.03$ MeV, $\displaystyle 7.07$ MeV and $\displaystyle 7.86$ MeV, identify the correct statement :
Official answer
(1)
energy of $\displaystyle 11.9$ MeV has to be supplied
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.