Question

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

• A. 8.3 MeV energy will be released
• B. energy of 12.4 MeV will be supplied
• C. energy of 3.6 MeV has to be supplied
• D. none of the above

Answer 1

Answer: (D)

Energy absorbed;

$\Delta Q=\text{Binding energy of products - Binding energy of reactants}$

$\text{Binding energy of products}$

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

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

$=150.88MeV$

$\text{Binding energy of Reactants}$ = $20\times 8.03=160.6MeV$

$\Delta Q=\text{Binding energy of products - Binding energy of reactants}=150.88-160.6=-9.72MeV$

$\text{Hence, 9.72 MeV energy will be released}$