Binding Energy and Nuclear Stability

The stability of a nucleus for a given mass number (A) is related to its binding energy. For naturally stable isobars (nuclei with the same mass number A but different atomic numbers Z), the nucleus will have the maximum possible binding energy. This corresponds to the minimum mass for a given A.

The given binding energy formula is the semi-empirical mass formula (excluding the pairing term, which is dependent on whether Z and N=A-Z are even or odd, but is not needed for finding the continuous dependence on Z):

\(B = a_1 A - a_2 A^{2/3} - a_3 Z^2 A^{-1/3} - a_4 (A - 2Z)^2 A^{-1}\)

Here, A is the mass number, Z is the atomic number, and \(a_1, a_2, a_3, a_4\) are coefficients. The stability condition for isobars (constant A) is found by maximizing the binding energy B with respect to Z. Mathematically, this means finding Z such that \(\frac{\partial B}{\partial Z} = 0\).

Calculating Z for Stable Isobars

Let's differentiate the binding energy formula with respect to Z, treating A as a constant:

\[\frac{\partial B}{\partial Z} = \frac{\partial}{\partial Z} (a_1 A) - \frac{\partial}{\partial Z} (a_2 A^{2/3}) - \frac{\partial}{\partial Z} (a_3 Z^2 A^{-1/3}) - \frac{\partial}{\partial Z} (a_4 (A - 2Z)^2 A^{-1})\]

• The first two terms, \(a_1 A\) and \(a_2 A^{2/3}\), do not depend on Z, so their derivatives with respect to Z are 0.
• For the third term, \(- a_3 Z^2 A^{-1/3}\), the derivative is \(- a_3 A^{-1/3} \cdot (2Z) = -2 a_3 Z A^{-1/3}\).
• For the fourth term, \(- a_4 (A - 2Z)^2 A^{-1}\), we use the chain rule. Let \(u = A - 2Z\). Then \(\frac{\partial u}{\partial Z} = -2\). The term is \(-a_4 u^2 A^{-1}\). The derivative is \(- a_4 A^{-1} \cdot \frac{\partial}{\partial Z}(A - 2Z)^2 = - a_4 A^{-1} \cdot 2(A - 2Z) \cdot (-2) = 4 a_4 A^{-1} (A - 2Z)\).

Setting the total derivative to zero for stability:

\[\frac{\partial B}{\partial Z} = -2 a_3 Z A^{-1/3} + 4 a_4 A^{-1} (A - 2Z) = 0\]

Rearranging the equation to solve for Z:

\[4 a_4 A^{-1} (A - 2Z) = 2 a_3 Z A^{-1/3}\]

Divide both sides by 2:

\[2 a_4 A^{-1} (A - 2Z) = a_3 Z A^{-1/3}\]

Expand the left side:

\[2 a_4 A^{-1} A - 2 a_4 A^{-1} (2Z) = a_3 Z A^{-1/3}\]

\[2 a_4 - 4 a_4 Z A^{-1} = a_3 Z A^{-1/3}\]

Move terms with Z to one side:

\[2 a_4 = a_3 Z A^{-1/3} + 4 a_4 Z A^{-1}\]

Factor out Z:

\[2 a_4 = Z (a_3 A^{-1/3} + 4 a_4 A^{-1})\]

Solve for Z:

\[Z = \frac{2 a_4}{a_3 A^{-1/3} + 4 a_4 A^{-1}}\]

To match the form of the options, let's divide the numerator and denominator by \(a_3\) and also multiply numerator and denominator by A:

\[Z = \frac{2 (a_4/a_3)}{(a_3/a_3) A^{-1/3} + 4 (a_4/a_3) A^{-1}}\]

\[Z = \frac{2 (a_4/a_3)}{A^{-1/3} + 4 (a_4/a_3) A^{-1}}\]

Multiply numerator and denominator by A:

\[Z = \frac{2 (a_4/a_3) A}{(A^{-1/3} + 4 (a_4/a_3) A^{-1}) A}\]

\[Z = \frac{2 (a_4/a_3) A}{A^{2/3} + 4 (a_4/a_3)}\]

We are given that \(\frac{a_4}{a_3} \simeq 30\). Substitute this value:

\[Z = \frac{2 \times 30 \times A}{A^{2/3} + 4 \times 30}\]

\[Z = \frac{60 A}{A^{2/3} + 120}\]

Result for Atomic Number Z

The atomic number Z for naturally stable isobars, given the relationship between the coefficients, is found to be:

\[Z = \frac{60 A}{120 + A^{2/3}}\]

Comparing this result with the given options, we find that it matches option 3.