Problem 21 - Fundamentals of Physics

Question

(a) Show that the total binding energy \(E_{\mathrm{bc}}\) of a given nuclide is $$E_{\mathrm{be}}=Z \Delta_{\mathrm{H}}+N \Delta_{\mathrm{n}}-\Delta$$ where \(\Delta_{\mathrm{H}}\) is the mass excess of \({ }^{1} \mathrm{H}, \Delta_{\mathrm{n}}\) is the mass excess of a neutron, and \(\Delta\) is the mass excess of the given nuclide. (b) Using this method, calculate the binding energy per nucleon for \({ }^{197}\) Au. Compare your result with the value listed in Table \(42-1 .\) The needed mass excesses, rounded to three significant figures, are \(\Delta_{\mathrm{H}}=+7.29 \mathrm{MeV}\) \(\Delta_{n}=+8.07 \mathrm{MeV},\) and \(\Delta_{197}=-31.2 \mathrm{MeV} .\) Note the economy of calculation that results when mass excesses are used in place of the actual masses.

Step-by-Step Solution

Answer

The binding energy per nucleon for \(^{197}\text{Au}\) is approximately \(7.91 \, \text{MeV/nucleon}\).

Step 1: Understanding the Formula

The formula given in the exercise is used to calculate the total binding energy \(E_{\mathrm{be}}\) of a nuclide. Binding energy is the energy required to disassemble a nucleus into its component protons and neutrons. \(Z\) and \(N\) represent the number of protons and neutrons, respectively. The formula is \(E_{\mathrm{be}}=Z \Delta_{\mathrm{H}}+N \Delta_{\mathrm{n}}-\Delta\). Here, \(\Delta_{\mathrm{H}}\) and \(\Delta_{\mathrm{n}}\) are the mass excesses of a proton (Hydrogen nucleus) and a neutron, respectively, and \(\Delta\) is the mass excess of the nuclide itself.

Step 2: Applying the Formula to Gold-197

\(^{197}\text{Au}\) has 79 protons and 197 - 79 = 118 neutrons. We use the formula: \(E_{\mathrm{be}}=79 \times 7.29 + 118 \times 8.07 - (-31.2)\). The binding energy \(E_{\mathrm{be}}\) thus computed gives us the total binding energy for \(^{197}\text{Au}\).

Step 3: Calculation for Total Binding Energy

Computing the values: \(79 \times 7.29 = 575.91\), \(118 \times 8.07 = 952.26\). Therefore, \(E_{\mathrm{be}} = 575.91 + 952.26 + 31.2\). The result is \(E_{\mathrm{be}} = 1559.37\, \text{MeV}\).

Step 4: Binding Energy per Nucleon

To find the binding energy per nucleon, divide the total binding energy by the number of nucleons: \(\frac{E_{\mathrm{be}}}{197} = \frac{1559.37}{197} \approx 7.91 \, \text{MeV/nucleon}\).

Step 5: Comparing with Table Value

The calculated binding energy per nucleon for \(^{197}\text{Au}\) is approximately \(7.91\, \text{MeV/nucleon}\). We compare this with the similar value found in Table 42-1 to ensure consistency and accuracy of our calculation.

Key Concepts

mass excess

Mass excess is a useful concept in nuclear physics that helps streamline calculations of nuclear properties. It represents the difference between the actual mass of a nuclide and its mass number when expressed in atomic mass units (amu). The mass excess is used rather than actual masses because it allows for easier computation of properties like binding energy.

Mass excess values are typically given in energy units, specifically mega-electronvolts (MeV), because of the mass-energy equivalence principle articulated by Einstein's famous equation, \(E=mc^2\). This allows us to directly find the energy-related properties of nuclei without the need for converting mass values directly. Let's look at some of its advantages:

• Facilitates quick calculations of binding energy by using a more direct subtraction operation.
• Reduces the potential for error compared to dealing with extremely small mass values in kilograms or amu.
• Allows physicists to use known values like the mass excess of a proton or a neutron to find unknown values for complex nuclei.

Understanding this concept is crucial for comprehensively analyzing nuclear reactions and the stability of various nuclides.

nucleons

Nucleons are the particles that make up an atomic nucleus: protons and neutrons. Each element is characterized by a unique number of protons, while the number of neutrons may vary, leading to different isotopes. The total number of nucleons in an atomic nucleus is called the mass number.

When calculating the binding energy of a nucleus, it's essential to know the number of protons (\(Z\)) and neutrons (\(N\)) since these values directly influence the total energy required to hold the nucleus together.

• Protons are positively charged particles that determine the chemical identity of the atom.
• Neutrons have no electrical charge but add to the mass and stability of the nucleus.
• The number and arrangement of nucleons affect the nuclear forces in the nucleus and thus its binding energy.

The binding energy per nucleon is a crucial measurement, giving insight into the stability of a nuclide. A higher binding energy per nucleon typically indicates a more stable nucleus.

nuclide

A nuclide is a distinct species of an atom characterized by a specific number of protons and neutrons in its nucleus. Each nuclide is defined by two quantities: its atomic number (\(Z\), the number of protons) and its mass number (\(A\), the total number of nucleons).

Nuclides are at the heart of nuclear chemistry and physics, and they vary significantly in their stability. Unstable nuclides are typically radioactive and will decay into more stable forms over time. Key points about nuclides include:

• Each nuclide is associated with a unique mass excess, used in its binding energy calculations.
• Nuclides with similar \(Z\) but different \(A\) values represent isotopes of an element.
• The study of nuclides, also known as nuclide chart analysis, helps scientists understand nuclear reactions, decay pathways, and fusion and fission processes.

With these fundamentals, students can better comprehend complex nuclear phenomena and calculation methods, especially those involving binding energies.