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Nuclear structure

An atomic nucleus is a quantum many-body system made of protons and neutrons, collectively called nucleons. It is not a rigid cluster of colored spheres at fixed positions. Different approximations reveal different features: a liquid drop captures broad binding trends, while a shell model explains unusually stable nucleon numbers and many spins and parities.

Naming a nucleus

Standard nuclide notation is

\[{}^A_ZX\]

where \(X\) is fixed by proton number \(Z\), neutron number is \(N\), and

\[A=Z+N\]

is mass number. Isotopes share \(Z\) but have different \(N\). Isotones share neutron number, and isobars share mass number.

Nuclear charge is \(+Ze\). Mass number is a count, not a mass measured in atomic mass units. Nuclear masses are not exactly \(A\) atomic mass units because binding changes the total rest energy.

def validate_nuclide(protons, neutrons, mass_number):
    if not all(isinstance(n, int) for n in [protons, neutrons, mass_number]):
        raise ValueError("Nucleon counts must be integers.")
    if protons < 0 or neutrons < 0 or mass_number <= 0:
        raise ValueError("Invalid nucleon counts.")
    if protons + neutrons != mass_number:
        raise ValueError("Mass number must equal protons plus neutrons.")
    return {"Z": protons, "N": neutrons, "A": mass_number}


print(validate_nuclide(2, 2, 4))

Nuclear size and density

Electron-scattering and nuclear-reaction measurements support the approximate radius relation

\[R=r_0A^{1/3}\]

with \(r_0\approx1.2\,\mathrm{fm}\). This is a characteristic radius for a diffuse quantum distribution, not a sharp hard-sphere boundary.

Because volume scales as \(R^3\),

\[V\approx\frac{4\pi}{3}r_0^3A\]

so mass density is roughly independent of \(A\) across many nuclei. This reflects the short range and saturation of the residual strong nuclear interaction: a nucleon interacts strongly with nearby nucleons rather than equally with every nucleon in the nucleus.

Proton electrostatic repulsion is long-range within the nucleus and grows in importance for large \(Z\). Competition among attraction, surface effects, Coulomb repulsion, and quantum statistics shapes stability.

Mass defect and binding energy

For bare nuclear masses,

\[\Delta m=Zm_p+Nm_n-m_{\mathrm{nucleus}}\]

and total binding energy is

\[B=\Delta mc^2\]

Positive \(B\) is the energy required to separate the ground-state nucleus into free protons and neutrons. The bound nucleus has less mass than those separated particles.

Tables usually provide neutral atomic masses. Electron masses cancel cleanly if hydrogen-atom mass \(m_{\mathrm H}\) is used:

\[B=\left[Zm_{\mathrm H}+Nm_n-m_{\mathrm{atom}}({}^A_ZX)\right]c^2\]

This ignores small differences in electronic binding energy at nuclear-energy precision. Do not mix this atomic-mass formula with \(m_p\) without also accounting for electrons.

Using \(1\,\mathrm u\,c^2\approx931.494\,\mathrm{MeV}\),

# Approximate neutral-atom masses in u for helium-4 binding energy.
hydrogen_u, neutron_u, helium_u = 1.007825, 1.008665, 4.002603
mass_defect_u = 2 * hydrogen_u + 2 * neutron_u - helium_u
binding_mev = mass_defect_u * 931.494
print(f"Binding energy: {binding_mev:.3f} MeV, per nucleon: {binding_mev/4:.3f} MeV")

Binding energy per nucleon

\[\frac{B}{A}\]

is an average comparison metric. It does not say every nucleon occupies the same state or requires the same removal energy.

Binding trend and stability

Binding energy per nucleon rises rapidly among light nuclei, reaches a broad maximum in the iron–nickel region, and declines gradually for very heavy nuclei. The trend helps explain why fusion of light nuclei and fission of heavy nuclei can release energy. It does not determine reaction rate, probability, or safety.

Stable light nuclei commonly have similar proton and neutron counts. In heavier nuclei, extra neutrons help supply attractive nuclear interaction without adding proton–proton Coulomb repulsion, so stable nuclei typically have \(N>Z\). Nuclei far from the valley of stability can lower energy through beta decay, particle emission, fission, or other transformations when allowed.

The semi-empirical mass formula

The liquid-drop model organizes broad binding trends as

\[B(A,Z)\approx a_vA-a_sA^{2/3}-a_c\frac{Z(Z-1)}{A^{1/3}}-a_a\frac{(A-2Z)^2}{A}+\delta(A,Z)\]

The terms represent:

The pairing term is typically positive for even–even nuclei, negative for odd–odd nuclei, and zero in a simple convention for odd \(A\). Coefficient values depend on the fit and convention. This formula predicts trends, not precision masses or detailed excited states.

A, Z = 56, 26
volume = 15.75 * A
surface = 17.8 * A ** (2 / 3)
coulomb = 0.711 * Z * (Z - 1) / A ** (1 / 3)
asymmetry = 23.7 * (A - 2 * Z) ** 2 / A
pairing = 12 / A**0.5  # Even-even nucleus; illustrative coefficient set.
binding = volume - surface - coulomb - asymmetry + pairing
print(f"Semi-empirical binding energy per nucleon: {binding/A:.3f} MeV")

Separation energies

The energy needed to remove one neutron is

\[S_n(A,Z)=B(A,Z)-B(A-1,Z)\]

and the proton separation energy is

\[S_p(A,Z)=B(A,Z)-B(A-1,Z-1)\]

These differences are not equal to \(B/A\). A nucleus can have a large average binding energy yet a weakly bound outer nucleon. When a separation energy becomes nonpositive, the corresponding particle-emission channel is energetically open, defining a drip-line boundary.

Nuclear shells

Like electrons in atoms, nucleons occupy quantum states. Protons and neutrons fill separate sets of nuclear levels subject to the Pauli principle. Strong spin–orbit coupling is essential to the observed shell ordering. Closed shells occur at the magic numbers

\[2,\ 8,\ 20,\ 28,\ 50,\ 82,\ 126\]

where 126 is most firmly established as a neutron shell closure. Nuclei with magic proton or neutron counts often show extra stability; doubly magic nuclei have both.

The shell model helps explain ground-state spin and parity, magnetic moments, and excitation patterns. A simple independent-particle shell picture still needs effective interactions and does not capture every collective rotation, vibration, or deformation.

Spin, parity, and deformation

Nuclear states are labeled by total angular momentum \(J\) and parity \(\pi=+1\) or \(\pi=-1\), often written \(J^\pi\). Even–even ground states commonly have \(J^\pi=0^+\) because nucleons tend to pair.

Many nuclei are not spherical. Quadrupole deformation can produce rotational bands, while vibrations of the nuclear surface create collective excitations. A level scheme therefore contains more information than a liquid-drop binding energy.

Energy scale and interaction range

Typical nuclear binding and excitation energies are measured in megaelectronvolts, far above ordinary atomic electronvolt scales. That difference is why a chemical reaction normally cannot rearrange a nucleus, even though electron and nuclear states belong to the same complete physical system.

The force binding nucleons is a residual effect of the strong interaction among quarks and gluons. At the nuclear-model scale it is short-ranged, has spin and isospin dependence, and includes a strongly repulsive behavior at very short distance. Calling it simply “attractive” misses the mechanism that prevents nuclei from collapsing into an arbitrarily dense point. Coulomb repulsion among protons remains important at the same time.

What to check

A reliable nuclear record should:

Nuclear structure ultimately reflects the strong, electromagnetic, and weak interactions in a quantum many-body system. The models here explain organization and energy scales but omit detailed nucleon–nucleon forces, three-body interactions, continuum coupling, and precision ab initio calculations.

A worked Python notebook

Convert a consistently defined mass defect into a nuclear binding energy. The question for this calculation is precise: how does binding energy change when we change mass defect, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

All masses must use the same atomic or nuclear convention before their difference is taken. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 0.05 u. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for mass defect, expressed in u. The output is binding energy, expressed in MeV. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 931.494 q\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 0.030377 u, the expression gives 28.296 MeV. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("931.494*q", locals={"q": q})
reference_input = 0.030377
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "u")
print("Predicted binding energy:", reference_output, "MeV")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 MeV. The midpoint, at 0.025 u, gives 23.2874 MeV. The upper endpoint gives 46.5747 MeV. Read these together with the assumptions: they describe different choices of mass defect within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 0.025, 0.05])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} u -> {output_value:.6g} MeV")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows mass defect; the vertical axis shows binding energy. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 0.05, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Mass defect (u)")
ax.set_ylabel("Binding energy (MeV)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does binding energy change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.030377 u is 931.494 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 931.494\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 0.030627
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "MeV")
print("Full expression:", full_prediction, "MeV")
print("Approximation difference:", full_prediction - linear_prediction, "MeV")

Here the changed input is 0.030627 u. The tangent estimate is 28.5289 MeV, while the full expression gives 28.5289 MeV. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 0.05
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of mass defect inside the stated interval, predict whether binding energy will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: All masses must use the same atomic or nuclear convention before their difference is taken. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of MeV: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(0.05) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 0.05, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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