What is NZ Graph? | Maximizing Binding Energy

New Section

This section introduces the concept of the NZ graph, which is a graphical representation of nucleus species based on the number of protons and neutrons. The stability curve represents nuclei with maximum stability and highest binding energy per nucleon.

The NZ Graph and Stability Curve

• The NZ graph represents all nucleus species in nature based on the number of protons and neutrons.

• The stability curve corresponds to nuclei at the center of the NZ graph, indicating maximum stability and high binding energy per nucleon.

• Nuclei that fall on the stability curve are more stable compared to those further away from it.

• Nuclei far from the stability curve may undergo radioactive decay processes like beta decay.

New Section

This section discusses two observations related to the NZ graph that provide insights into nuclear stability. It also explains how smaller nuclei tend to have equal numbers of neutrons and protons, while larger nuclei have an excess of neutrons.

Observations from the NZ Graph

• The N is equal to Z line represents points where neutron and proton numbers are exactly equal.

• For smaller nuclei (mass number less than 20), the NZ graph almost coincides with the N is equal to Z line, indicating that equal numbers of neutrons and protons lead to greater stability.

• As mass number increases, there is an excess of neutrons compared to protons in stable configurations.

New Section

This section explains how neutron and proton organization in distinct energy levels contributes to nuclear stability. It also discusses how configurations with an excess number of neutrons have higher total energy, leading to a tendency for systems to evolve towards configurations with lower energy.

Neutron-Proton Energy Levels

• Neutrons and protons organize themselves in distinct energy levels within the nucleus.

• No more than two neutrons or protons can occupy the same energy level due to the exclusion principle.

• Configurations with an excess of neutrons have higher total energy compared to configurations with equal numbers of neutrons and protons.

• Natural systems tend to evolve towards configurations with lower total energy, leading to radioactive decay processes like beta decay.

New Section

This section explores how beta decay processes can decrease the overall energy of a configuration by converting excess neutrons into protons. It also highlights the stability and transformation of specific nucleus species.

Beta Decay and Stability

• In beta decay, a neutron can convert into a proton or vice versa, resulting in a more stable configuration.

• When an excess neutron converts to a proton, the proton occupies an energy state lower than its corresponding neutron counterpart.

• The overall energy of the configuration decreases through beta decay, making it more stable.

• Certain nucleus species undergo beta decay processes to become more stable versions.

The transcript is already in English.

Nuclear Forces and Nucleus Size

This section discusses the forces that exist inside the nucleus and how they affect the stability of nuclei as their size increases.

Forces Inside the Nucleus

• The nucleus is held together by two types of forces: the strong nuclear force and the Coulomb repulsion.

• The strong nuclear force acts between all particles in the nucleus and is attractive in nature, while the Coulomb repulsion only exists between protons and tries to break apart the nucleus.

• At short distances, the strong nuclear force dominates over the Coulomb repulsion, resulting in stable nuclei with small sizes.

• However, as nuclei become larger, there is no strong nuclear attraction between particles at opposite ends of the nucleus due to increased distances. The Coulomb repulsion can still act between nearest neighbor protons and those at opposite ends.

Impact on Nucleus Stability

• As nucleus size increases, the Coulomb repulsion starts to dominate over the strong nuclear force.

• To maintain a stable nuclear structure, an excess number of neutrons is needed to compensate for this increasing Coulomb repulsion.

• Excess neutrons contribute to stronger nuclear forces, while excess protons would lead to more coulombic repulsion.

• Therefore, larger nuclei require an excess number of neutrons compared to protons for stability.

Theoretical Prediction of NZ Graph

This section explores whether it is possible to theoretically predict the stability curve shown in an NZ graph.

Using Binding Energy Expression

• The binding energy expression obtained from a semi empirical binding energy formula can be used for theoretical predictions.

• By inputting a fixed mass number (A), it is possible to obtain combinations of neutrons (N) and protons (Z) that result in stable nuclear configurations.

• Most combinations will have low binding energies and are not stable, while only a few combinations will yield high binding energies and stability.

• The value of Z for which the binding energy is maximum corresponds to the most stable version of the nucleus.

Maximizing Binding Energy

• To find the value of Z that maximizes the binding energy, we can take the derivative of the binding energy expression with respect to Z and set it equal to zero.

• After performing the derivative, terms involving Z in the third and fourth terms need to be considered.

• By solving this equation, we can determine the value of Z that corresponds to the most stable configuration for a given mass number.

For more detailed explanations on nuclear forces, nucleus size, and theoretical predictions using binding energy expressions, refer to specific sections in previous videos.

Simplifying the Expression

In this section, the speaker simplifies an expression and brings it to the left-hand side.

Simplification Steps

• The expression can be written as 4n^4A(A-2Z) = A^(3/2)(Z-1)/A^(1/3).

• Bring the expression to the left-hand side: (3/A^4)(A/A^(1/3)) = (4A-2Z)/(2Z-1).

Approximation of the Expression

The speaker approximates the simplified expression.

Approximation Steps

• The expression can be approximated as (1/2)(A^(3/4))(A^(2/3)) ≈ (A-2Z)/Z.

Further Approximation

The speaker makes another approximation based on large number values.

Further Approximation Steps

• Since A and Z have large number values, -1/2 can be approximated as Z.

• Therefore, the expression becomes n + Z - 2Z/Z ≈ (1/2)(A^(3/4))(A^(2/3)).

• Simplifying further, n/Z - 1 ≈ (1/2)(A^(3/4))(A^(2/3)).

• Rearranging, n/Z ≈ (1/2)(A^(3/4))(A^(2/3)) + 1.

Obtaining a Mathematical Function for NZ Graph

The speaker derives a mathematical function that approximates the stability curve in the NZ graph.

Derivation Steps

• Since A = n + Z, the expression can be written as n + Z - 2Z/Z ≈ (1/2)(A^(3/4))(A^(2/3)) + 1.

• Simplifying further, n/Z - 1 ≈ (1/2)(A^(3/4))(A^(2/3)) + 1.

• Rearranging, n/Z ≈ (1/2)(A^(3/4))(A^(2/3)) + 1.

• The obtained expression represents the stability curve or the most stable version of a given nucleus.

• This expression can be used to approximate the stability curve and plot it on an NZ graph.

Values of Constants A3 and A4

The speaker discusses the values of constants A3 and A4.

Constant Values

• The values of A3 and A4 are specific to Coulomb big repulsive energy dome and symmetric energy term respectively.

Plotting the Stability Curve

The speaker demonstrates a program in Scilab to plot the stability curve based on the derived mathematical function.

Plotting Steps

• By running a program in Scilab, the N = Z line is plotted first.

• Then, the stability curve corresponding to the NZ graph is plotted.

• For small nuclei sizes, the stability curve coincides with N = Z line. As mass number increases, neutrons start exceeding protons.

Answering an Original Question

The speaker answers a question about determining the most stable configuration for a given mass number.

Determining Most Stable Configuration

• For a given mass number (e.g., A = 25), the most stable configuration can be determined using the derived expression for n/Z.

• By substituting A = 25 into the expression, the value of Z is calculated as approximately 11.71.

• Rounding off to a whole number, Z ≈ 12.

• Therefore, for A = 25, the most stable configuration is with 12 protons and (25 - 12) = 13 neutrons.

Configuration of Neutrons and Protons

The speaker explains how to obtain the neutron-proton configuration corresponding to maximum stability for any given mass number.

Obtaining Neutron-Proton Configuration

• The neutron-proton ratio that corresponds to maximum stability for a given mass number can be obtained using the derived expression for n/Z.

• This calculation can be done for any other given mass number.