Q. Condiser the nuclear fission $\ce{Ne^{20} -> 2 He^{4} + C^{12}} $
Given that the binding energy/nucleon of $\ce{Ne^{20}, He^{4}}$ and $\ce{C^{12}}$ are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :

Solution:

$\ce{Ne^{20} -> 2He^{4} + C^{12}}$
$8.30 \times 20 \; \; \; \; 2\times 7.07 \times 4 + 7.86 \times 12$
$\therefore \; E_B = (BE)_{react} - (BE)_{product} = 9.72 \, MeV$

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