Question

The binding energy B(A,Z) of an atomic nucleus of mass number A, atomic number Z, and number of neutrons N = A-Z, can be expressed as \[ B(A,Z) = a_1 A - a_2 A^{2/3} - a_3 \frac{Z^2}{A^{1/3}} - a_4 \frac{(A-2Z)^2}{A} \] where \(a_1, a_2, a_3\), and \(a_4\) are constants of appropriate dimensions. Let \(B(A, Z')\) be the binding energy of a mirror nucleus (which has the same A, but the number of protons and neutrons are interchanged).
Then, at constant A, \([B(A,Z) - B(A,Z')]\) is

• A. proportional to \(Z^2\)
• B. proportional to \((Z^2 - N^2)\)
• C. proportional to \(N^2\)
• D. constant

Hint:

For mirror nuclei, the only term in the SEMF (as given here) that differs is the Coulomb term, because the number of protons (Z) changes. The asymmetry term, which depends on \((N-Z)^2\), remains the same because \(|N'-Z'| = |Z-N| = |N-Z|\). This simplifies the comparison significantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question uses the Semi-Empirical Mass Formula (SEMF) to compare the binding energies of mirror nuclei. Mirror nuclei are pairs of isobars (same mass number A) where the proton number of one equals the neutron number of the other, and vice versa. The difference in their binding energies primarily arises from the Coulomb term, as protons experience electrostatic repulsion while neutrons do not.
Step 2: Key Formula or Approach:
We are given the binding energy formula for a nucleus (A, Z):
\[ B(A,Z) = a_1 A - a_2 A^{2/3} - a_3 \frac{Z^2}{A^{1/3}} - a_4 \frac{(A-2Z)^2}{A} \] For its mirror nucleus (A, Z'), the number of protons is \(Z' = N\) and the number of neutrons is \(N' = Z\), where \(N = A-Z\). We need to write the expression for \(B(A, Z')\) and then calculate the difference \([B(A,Z) - B(A,Z')]\).
Step 3: Detailed Explanation:
Let's analyze the terms of the SEMF for the mirror nucleus \((A, Z')\).
The proton number is \(Z' = N = A-Z\).
The binding energy \(B(A, Z')\) is: \[ B(A, Z') = a_1 A - a_2 A^{2/3} - a_3 \frac{(Z')^2}{A^{1/3}} - a_4 \frac{(A-2Z')^2}{A} \] Let's examine the asymmetry term for the mirror nucleus:
The term \((A-2Z')^2\) becomes \((A - 2N)^2 = (A - 2(A-Z))^2 = (A - 2A + 2Z)^2 = (2Z-A)^2 = (A-2Z)^2\).
So, the asymmetry term is identical for both nuclei. The volume term (\(a_1 A\)) and surface term (\(a_2 A^{2/3}\)) also only depend on A, so they are the same.
Therefore, when we take the difference \(B(A,Z) - B(A,Z')\), the volume, surface, and asymmetry terms will cancel out.
\[ B(A,Z) - B(A,Z') = \left(-a_3 \frac{Z^2}{A^{1/3}}\right) - \left(-a_3 \frac{(Z')^2}{A^{1/3}}\right) \] Substitute \(Z' = N\):
\[ B(A,Z) - B(A,Z') = -a_3 \frac{Z^2}{A^{1/3}} + a_3 \frac{N^2}{A^{1/3}} \] Factor out the common terms:
\[ B(A,Z) - B(A,Z') = \frac{a_3}{A^{1/3}} (N^2 - Z^2) \] Since A and \(a_3\) are constants for this comparison, the difference in binding energy is directly proportional to \((N^2 - Z^2)\).
The expression in option (B) is \((Z^2 - N^2)\). Since \((N^2 - Z^2) = -1 \times (Z^2 - N^2)\), being proportional to \((N^2 - Z^2)\) is equivalent to being proportional to \((Z^2 - N^2)\).
Step 4: Final Answer:
The difference \([B(A,Z) - B(A,Z')]\) is proportional to \((N^2 - Z^2)\), which is equivalent to being proportional to \((Z^2 - N^2)\). This corresponds to option (B).