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Consider the nuclear fission $N e^{20} \to 2 H e^{4} + C^{12}$ . Given that the binding energy per nucleon of $N e^{20}, H e^{4}$ and $C^{12}$ are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :

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Energy of 9.72 MeV has to be supplied

8.3 MeV energy will be released

Energy of 3.6 MeV will be released

Energy of 12.4 MeV will be supplied

answer is D.

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Detailed Solution

$T . E_{Re a c t i o n s} = 20 (8.03) = 160.6$
$T . E_{p r o d u c t s} = 8 (7.07) + 12 (7.86)$
$= 56.56 + 94.32$
$= 150.88$
$∵ T . E_{r e a c t i o n} > T . E_{p r o d u c t i o n}$ energy is absorbed, so 9.72 MeV has to be supplied

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