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Nuclear energy

Nuclear reactions rearrange nucleons and nuclear states. Because rest energy is part of total energy, a small difference between initial and final rest masses can appear as kinetic energy, radiation, or excitation. Energy accounting begins with a complete balanced reaction, not with an image of an explosion.

A positive energy release says nothing by itself about reaction probability, rate, controllability, engineering efficiency, radiation transport, or safety. Those require additional physics and system information.

Reaction bookkeeping

A generic nuclear reaction can be written

\[a+A\rightarrow B+b+\cdots\]

Ordinary nuclear reactions conserve electric charge and total energy–momentum. Mass number is also conserved when only nucleons and nuclei are rearranged:

\[\sum_{\mathrm{initial}}Z=\sum_{\mathrm{final}}Z,\qquad \sum_{\mathrm{initial}}A=\sum_{\mathrm{final}}A\]

Beta processes require leptons and conserve lepton number in the Standard Model. Mass number is useful nuclear bookkeeping, not a universal fundamental quantum number for every high-energy particle reaction.

# Deuterium + tritium -> helium-4 + neutron; entries are (Z, A).
reactants = [(1, 2), (1, 3)]
products = [(2, 4), (0, 1)]
assert sum(z for z, a in reactants) == sum(z for z, a in products)
assert sum(a for z, a in reactants) == sum(a for z, a in products)
print("Charge and nucleon bookkeeping passed; kinematics still needs checking.")

Balancing charge and nucleon count is necessary, but not sufficient. Angular momentum, parity, lepton number, and other quantum numbers can forbid or suppress a channel.

The Q value

Using compatible rest masses,

\[Q=\left(\sum_i m_{i,\mathrm{initial}}-\sum_fm_{f,\mathrm{final}}\right)c^2\]

If \(Q>0\), the reaction is exoergic and can release that rest-energy difference as product kinetic energy or radiation. If \(Q<0\), at least \(|Q|\) must be supplied, and momentum conservation can make the laboratory threshold larger.

When masses are in atomic mass units,

\[1\,\mathrm u\,c^2\approx931.494\,\mathrm{MeV}\]

provides the conversion. Neutral atomic masses can simplify electron bookkeeping only when electron counts cancel appropriately. Positron emission and electron capture require special care. Never mix atomic and bare nuclear masses silently.

# Neutral atomic masses; electron counts match on both sides.
deuterium_u, tritium_u = 2.014102, 3.016049
helium_u, neutron_u = 4.002603, 1.008665
q_mev = (deuterium_u + tritium_u - helium_u - neutron_u) * 931.494
print(f"Approximate D-T reaction Q value: {q_mev:.3f} MeV")

Energy and momentum determine product energies

The Q value is not automatically the kinetic energy of one selected product. Total relativistic energy and vector momentum must both be conserved:

\[\sum_iE_i=\sum_fE_f,\qquad \sum_i\mathbf p_i=\sum_f\mathbf p_f\]

For a two-body final state in the center-of-momentum frame, the products have equal and opposite momenta. Their kinetic energies divide according to their masses. In a three-body beta decay, available energy can be shared continuously, which is why emitted beta particles have a spectrum rather than one energy.

An endoergic reaction initiated by a projectile striking a stationary target needs extra energy so the final center of mass can carry momentum. A nonrelativistic threshold approximation for

\(a+A\to B+b\)

is

\[K_{a,\mathrm{threshold}}\approx -Q\left(1+\frac{m_a}{m_A}\right)\]

when \(Q<0\) and mass changes are small. The exact threshold follows relativistic invariants.

Binding energy explains broad trends

Binding energy per nucleon rises steeply for light nuclei, has a broad maximum near the iron–nickel region, and slowly declines for heavy nuclei. Reactions that move nuclei toward more strongly bound configurations can release energy.

This explains two broad possibilities:

The curve does not say every fusion or fission channel occurs spontaneously. Energy barriers, quantum selection rules, cross sections, and available reaction pathways determine rates.

Fission energy

In fission, a heavy nucleus divides into two main fragments plus neutrons and often prompt gamma rays. A schematic neutron-induced event is

\[n+{}^A_ZX\rightarrow{}^{A_1}_{Z_1}Y+{}^{A_2}_{Z_2}Y'+\nu n+\gamma+\cdots\]

The fragment distribution is not one fixed pair. Energy appears mainly as fragment kinetic energy, with prompt and delayed radiation and neutrino energy also contributing to the complete ledger.

Some emitted neutrons can induce later fissions. A population-level chain reaction is often summarized by an effective multiplication factor \(k_{\mathrm{eff}}\):

This classification does not calculate a reactor. Geometry, neutron energies, materials, delayed neutrons, heat removal, feedback, and engineered control and safety systems are essential.

Fusion energy and the Coulomb barrier

Fusion joins light nuclei into more strongly bound products. Positively charged nuclei repel through the Coulomb interaction:

\[V(r)=\frac{Z_1Z_2e^2}{4\pi\epsilon_0r}\]

At stellar temperatures, average thermal kinetic energy is often below the classical barrier at nuclear distances. Quantum tunneling permits a small fraction of encounters to react. A positive Q value therefore does not imply a fast room-temperature reaction.

Reaction rates depend on particle distributions and energy-dependent cross sections. For two species with number densities \(n_1\) and \(n_2\), a thermal reaction rate per volume is

\[r=\frac{n_1n_2}{1+\delta_{12}}\langle\sigma v_{\mathrm{rel}}\rangle\]

The denominator prevents double-counting identical reactants.

\(\langle\sigma v_{\mathrm{rel}}\rangle\)

depends on the velocity distribution and nuclear cross section.

Cross section and reaction rate

A cross section \(\sigma(E)\) has units of area and quantifies reaction probability for an incident flux; it is not a physical geometric target size in every detail. For a thin target with number density \(n_t\), thickness \(x\), and incident particle rate \(\dot N_b\), a low-attenuation estimate is

\[\dot N_{\mathrm{reaction}}\approx\dot N_b\,n_tx\,\sigma\]

This assumes a sufficiently thin target, one relevant energy, and independent interactions. Thick targets require energy loss, attenuation, and secondary production.

From energy per reaction to power

If reactions occur at mean rate \(\dot N\) and deposit usable energy \(Q_{\mathrm{dep}}\) per reaction, thermal power is

\[P_{\mathrm{thermal}}=\dot NQ_{\mathrm{dep}}\]

The deposited energy can be smaller than the total Q value when neutrinos or escaping radiation carry energy away. Electrical output is further reduced by thermodynamic conversion and plant losses:

\[P_{\mathrm{electric}}=\eta P_{\mathrm{thermal}}\]

with \(0\leq\eta\leq1\). This is energy accounting, not a design model.

q_j = q_mev * 1e6 * 1.602176634e-19
reaction_rate, deposited_fraction, efficiency = 1e15, 0.8, 0.35
deposited_power = reaction_rate * q_j * deposited_fraction
electric_power = deposited_power * efficiency
print(
    f"Deposited power: {deposited_power:.3f} W; electric output: {electric_power:.3f} W"
)

Conservation and uncertainty checks

Before reporting a reaction result:

The Q-value uncertainty follows from mass uncertainties. Correlated mass measurements should not be combined as if independent. Rounding masses too early is dangerous because Q is a small difference between much larger rest masses.

This page intentionally does not provide reactor construction, weapon design, fuel-cycle operations, shielding dimensions, or medical dose guidance. Those are specialized high-stakes engineering domains. The equations here establish the physical energy ledger and the distinctions needed to interpret it.

A worked Python notebook

Assume 17.6 megaelectronvolts per reaction and a deposited fraction of 0.8. The question for this calculation is precise: how does deposited thermal power change when we change reaction rate, 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.

This accounting example does not model reaction kinetics, engineering feasibility, or a power plant. 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 1e+12 to 1e+15 1/s. 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 reaction rate, expressed in 1/s. The output is deposited thermal power, expressed in W. 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.

\[f(q) = 2.255864700672 \cdot 10^{-12} q\]

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 1e+14 1/s, the expression gives 225.586 W. 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("17.6e6*1.602176634e-19*0.8*q", locals={"q": q})
reference_input = 100000000000000.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "1/s")
print("Predicted deposited thermal power:", reference_output, "W")

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 2.25586 W. The midpoint, at 5.005e+14 1/s, gives 1129.06 W. The upper endpoint gives 2255.86 W. Read these together with the assumptions: they describe different choices of reaction rate 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([1000000000000.0, 500500000000000.0, 1000000000000000.0])
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} 1/s -> {output_value:.6g} W")
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 reaction rate; the vertical axis shows deposited thermal power. 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(1000000000000.0, 1000000000000000.0, 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("Reaction rate (1/s)")
ax.set_ylabel("Deposited thermal power (W)")
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 deposited thermal power change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1e+14 1/s is 2.25586e-12 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.

\[f'(q) = 2.255864700672 \cdot 10^{-12}\]

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 = 104995000000000.0
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, "W")
print("Full expression:", full_prediction, "W")
print("Approximation difference:", full_prediction - linear_prediction, "W")

Here the changed input is 1.04995e+14 1/s. The tangent estimate is 236.855 W, while the full expression gives 236.855 W. 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 = 999000000000000.0
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 reaction rate inside the stated interval, predict whether deposited thermal power 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: This accounting example does not model reaction kinetics, engineering feasibility, or a power plant. 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 W: 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(1000000000000000.0) - evaluate(1000000000000.0))
for points in [101, 201, 401]:
    grid = np.linspace(1000000000000.0, 1000000000000000.0, 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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