Question

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.

Short answer

Question: Show that the Bethe-Weizsäcker formula for odd A nuclei is a quadratic in Z and find the most stable isotope with A = 117. Answer: When considering the Bethe-Weizsäcker formula for odd A nuclei, we find that the expression becomes a quadratic in Z after substituting Z = A - N, where N is the number of neutrons. To find the most stable isotope with A = 117, we need to solve the expression for the maximum binding energy using the given constants, resulting in Zmax = (1/2)(117 - (aC/2aA)117^(-2/3)). Depending on the values of aC and aA, we can determine the most stable isotope with A = 117.

Step-by-step solution

Write the Bethe-Weizsäcker formula

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 Bethe-Weizsäcker formula 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 \(Z\).

Find the most deeply bound isotope with \(A = 117\)

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 \(A = 117\).

Key concepts

Nuclear Physics

Nuclear physics is the branch of physics that delves into the components and behavior of atomic nuclei. The fundamental forces active within a nucleus are pivotal. They determine how particles like protons and neutrons are bound together. Understanding these interactions helps explain why some nuclei are stable, while others are radioactive.

One significant milestone in nuclear physics was the development of the Bethe-Weizsäcker formula. This equation estimates the binding energy of a nucleus, a crucial factor in determining stability. Within a nucleus, protons and neutrons are held together by the strong nuclear force, which is considerably stronger than the electromagnetic force that tries to push the protons apart due to their positive charges. Thus, exploring these forces and behaviors gives us a deeper understanding of nuclear phenomena, including fission, fusion, and decay processes.

Binding Energy

Binding energy is a measure of the stability of a nucleus. It is the energy required to disassemble a nucleus into its separate protons and neutrons. A higher binding energy indicates a more stable nucleus.

In the context of the Bethe-Weizsäcker formula, the binding energy is calculated using several terms that correspond to various forces and interactions within the nucleus:

• The volume term accounts for the strong force binding the nucleons together.
• The surface term considers that nucleons on the surface have fewer neighboring nucleons, slightly reducing the binding energy.
• The Coulomb term reflects the repulsive force between protons, which decreases the binding energy.
• The asymmetry term adjusts for the proportion of protons to neutrons, penalizing deviations from equal numbers.
• For even-even nuclei, an additional term accounts for pairing effects, though it's zero for odd-A nuclei.

Binding energy is crucial because it influences nuclear processes and reactions. The difference in binding energy between reactants and products often appears as energy released in reactions, such as in nuclear power generation.

Isotopes

Isotopes are variations of elements that share the same number of protons but differ in the number of neutrons. This variation impacts the atomic mass and, consequently, the nuclear properties of the isotopes.

For instance, isotopes play a significant role in nuclear stability and radioactivity. Because isotopes of the same element have different neutron counts, their nuclear properties, including binding energy, can vary significantly. Isotopes with higher binding energy tend to be more stable. Understanding isotopes is crucial for applications ranging from nuclear medicine to carbon dating.

In the context of the Bethe-Weizsäcker formula, it gives insight into which isotopes of a given mass number (like 117 in this exercise) are the most stable. By analyzing isotopic variations, we can determine the most stable configuration, which is vital for nuclear research and applications.

Quadratic Expression in Z

The Bethe-Weizsäcker formula allows us to express binding energy as a quadratic function of the atomic number, \(Z\). This is crucial because it makes it easier to determine which isotope of a specific mass number \(A\) is the most stable.

Mathematically, a quadratic expression in \(Z\) takes the form \(a + bZ + cZ^2\). Here, \(a\), \(b\), and \(c\) are coefficients derived from simplifying the Bethe-Weizsäcker formula, considering the mass number \(A\) and constants related to nuclear properties. By setting the derivative of this expression to zero, one can determine \(Z_{max}\), the atomic number that maximizes the binding energy.

• Using this approach, the most stable isotope for a given \(A\) can be precisely identified, which is essential in the study of isotope stability and nuclear structure.
• It simplifies calculations and helps predict nuclear behavior without extensive computational methods.

This powerful tool underscores the relationship between nuclear physics and algebra, providing critical insight into nuclear stability.