Nuclei \(A\) and \(B\) form a nucleus \(C\). The binding energy per nucleon (BE/N) for \(A\), \(B\), and \(C\) are \(3\,\text{MeV}\), \(7\,\text{MeV}\), and \(6\,\text{MeV}\) respectively. Find the energy produced in the reaction:
\[
2A^3 + B^4 \rightarrow C^{10}
\]
Show Hint
Energy released in nuclear reactions is determined from the \textbf{change in total binding energy}.
If the final nucleus has higher binding energy, the excess energy is released.
Concept:
The energy released in a nuclear reaction equals the increase in the total binding energy of the system.
\[
\text{Energy released} = BE_{\text{final}} - BE_{\text{initial}}
\]
Also,
\[
\text{Total Binding Energy} = (\text{Binding Energy per nucleon}) \times (\text{Number of nucleons})
\]
Step 1: Calculate total binding energy of reactants.}
For nucleus \(A^3\):
\[
BE/N = 3\,\text{MeV}, \quad \text{nucleons} = 3
\]
\[
BE_A = 3 \times 3 = 9\,\text{MeV}
\]
Since there are two nuclei of \(A\):
\[
BE_{2A} = 2 \times 9 = 18\,\text{MeV}
\]
For nucleus \(B^4\):
\[
BE/N = 7\,\text{MeV}
\]
\[
BE_B = 7 \times 4 = 28\,\text{MeV}
\]
Total initial binding energy:
\[
BE_{\text{initial}} = 18 + 28 = 46\,\text{MeV}
\]
Step 2: Calculate total binding energy of product nucleus.}
For nucleus \(C^{10}\):
\[
BE/N = 6\,\text{MeV}
\]
\[
BE_C = 6 \times 10
\]
\[
BE_C = 60\,\text{MeV}
\]
Step 3: Find the energy released.}
\[
E = BE_{\text{final}} - BE_{\text{initial}}
\]
\[
E = 60 - 46
\]
\[
E = 14\,\text{MeV}
\]
