Consider the Bethe-Weizsäcker formula for the case of odd \(A\) nuclei. Show that the formula can be written as a quadratic in \(Z\) -and thus, that for any given \(A\), the binding energies of the isotopes having that \(A\) take a quadratic form, \(B=a+b Z+c Z^{2} .\) Find the most deeply bound isotope (the most stable one) having \(A=117\) using your result.

The Bethe-Weizsäcker formula for the binding energy of a nucleus with mass number \(A\) and atomic number \(Z\) is given by: \(B(Z, A) = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z - 1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^2}{A} - \delta(Z, A)\) Here, \(a_{V}, a_{S}, a_{C}, a_{A}\), and \(\delta(Z, A)\) are constants that depend on the isospin of the nuclei. For odd \(A\) nuclei, \(\delta(Z, A) = 0\). Now, let's rewrite the formula for binding energy for odd \(A\) nuclei.

Simplify the formula by removing the \(\delta(Z, A)\) term for odd \(A\) and rearrange the terms: \(B(Z, A) = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z - 1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^2}{A}\) Now, let \(N = A - Z\), where \(N\) is the number of neutrons. We can write \(Z\) in terms of \(A\) and \(N\): \(Z = A - N\) Substitute this into the Bethe-Weizsäcker formula: \(B(Z, A) = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{(A - N)(A - N - 1)}{A^{1/3}} - a_{A} \frac{(A - 2(A - N))^2}{A}\) Now, it becomes clear that \(B(Z, A)\) is a quadratic in \(Z\).

To find the most deeply bound isotope with \(A = 117\), we want to find the maximum binding energy for the isotopes. This means that we need to find when the derivative of \(B(Z, A)\) with respect to \(Z\) is equal to \(0\): \(\frac{dB(Z, A)}{dZ} = 0\) By solving this expression for \(Z\), we will find the \(Z\) value that gives the maximum binding energy for the isotopes with \(A = 117\). Differentiate \(B(Z, A)\) with respect to \(Z\) and set it to \(0\): \(\frac{d}{dZ}(a_{V} A - a_{S} A^{2/3} - a_{C} \frac{(A - N)(A - N - 1)}{A^{1/3}} - a_{A} \frac{(A - 2(A - N))^2}{A}) = 0\) To simplify calculations, one usually assumes that the binding energy per nucleon is constant for all nuclei. In this case, \(a_V = k\). So the quadratic expression for the binding energy becomes: \(B(Z, A) = k A - a_{S} A^{2/3} - a_{C} \frac{Z(Z - 1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^2}{A}\) Now, differentiate this quadratic equation to find the maximum binding energy: \(\frac{d}{dZ}(k A - a_{S} A^{2/3} - a_{C} \frac{Z(Z - 1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^2}{A}) = 0\) Using some algebraic manipulation and solving for \(Z = Z_{max}\): \(Z_{max} = \frac{1}{2} \left( A - \frac{a_C}{2a_A} A^{-2/3} \right)\) Now, insert \(A = 117\) to find the most stable isotope: \(Z_{max} = \frac{1}{2} \left( 117 - \frac{a_C}{2a_A} 117^{-2/3} \right)\) Depending on the values of \(a_C\) and \(a_A\), we can determine the most stable isotope with \(A = 117\).