Consider the nuclear fission, $N e^{20} \to 2 H e^{4} + C^{12}$
Given that the binding energy/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:

Question

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Answer 1

Energy absorbed;
$Δ Q = Binding energy of products - Binding energy of reactants$

$Binding energy of products$

$= 2 \times (B.E of H e^{4}) + (B.E of C^{12})$

$= 2 (4 \times 7.07) + (12 \times 7.86)$
$= 150.88 M e V$

$Binding energy of Reactants$ = $20 \times 8.03 = 160.6 M e V$

$Δ Q = Binding energy of products - Binding energy of reactants = 150.88 - 160.6 = - 9.72 M e V$

$Hence, 9.72 MeV energy will be released$

Answer 2

Energy absorbed;

Δ Q = Binding energy of products - Binding energy of reactants

Binding energy of products

= 2 \times (B.E of H e^{4}) + (B.E of C^{12})

= 2 (4 \times 7.07) + (12 \times 7.86)

= 150.88 M e V

Binding energy of Reactants

=

20 \times 8.03 = 160.6 M e V

Δ Q = Binding energy of products - Binding energy of reactants = 150.88 - 160.6 = - 9.72 M e V

Hence, 9.72 MeV energy will be released