Question

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: \[ 2A^3 + B^4 \rightarrow C^{10} \]

• A. \(8\,\text{MeV}\)
• B. \(12\,\text{MeV}\)
• C. \(14\,\text{MeV}\)
• D. \(10\,\text{MeV}\)

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.

Answer 1

The Correct Option is B

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.} \[ E = BE_{\text{final}} - BE_{\text{initial}} \] \[ E = 60 - 46 \] \[ E = 14\,\text{MeV} \]