Question

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process, \(\Delta Mc^2 =\) ______ MeV.

• A. 6.9
• B. 7.9
• C. 2.2
• D. 4.3

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a nuclear reaction, the \(Q\)-value or energy equivalence of the mass defect (\(\Delta Mc^2\)) can be found by comparing the total binding energy of the reactants with the total binding energy of the products.
Step 2: Key Formula or Approach:
The total binding energy (B.E.) of a nucleus is given by:
\(\text{Total B.E.} = \text{Mass Number } (A) \times \text{Binding Energy per nucleon}\).
The energy equivalent of the mass difference is:
\(|\Delta M c^2| = |\text{Total B.E. of products} - \text{Total B.E. of reactants}|\).
Step 3: Detailed Explanation:
The reaction involves two nuclei of \(A=3\) and one nucleus of \(A=4\) forming one nucleus of \(A=10\).
Total Binding Energy of reactants:
\(E_{\text{reactants}} = 2 \times (3 \times 5.6) + 1 \times (4 \times 7.4)\)
\(E_{\text{reactants}} = 2 \times 16.8 + 29.6 = 33.6 + 29.6 = 63.2 \text{ MeV}\).
Total Binding Energy of product:
\(E_{\text{products}} = 10 \times 6.1 = 61.0 \text{ MeV}\).
Now find the absolute difference in energy (which corresponds to \(\Delta Mc^2\)):
\(\Delta Mc^2 = |E_{\text{products}} - E_{\text{reactants}}|\)
\(\Delta Mc^2 = |61.0 - 63.2| = |-2.2| = 2.2 \text{ MeV}\).
Step 4: Final Answer:
The value of \(\Delta Mc^2\) is \(2.2\) MeV.