15.7 Fission, Fusion, and Nuclear Decay

Fission, fusion, and radioactive decay are nuclear processes controlled by the strong force, conservation laws, and mass-energy equivalence ($E = mc^2$). Fusion combines small nuclei, fission splits heavy nuclei, and radioactive decay shrinks a sample of unstable nuclei exponentially over time.

Why This Matters for the AP Physics 2 Exam

This topic shows up in both multiple-choice and free-response settings, and it pairs naturally with the kind of reasoning the Mathematical Routines free-response question rewards. You may be asked to balance a nuclear reaction, calculate energy released from a mass change, or derive how long it takes a sample to decay to a given fraction.

The strongest answers connect a formula to its physical meaning. Knowing that $N = N_0 e^{-\lambda t}$ describes exponential decay is good, but explaining why equal time intervals remove equal fractions (not equal numbers) of nuclei is what earns full justification credit. Expect to apply conservation of nucleon number, charge, energy, and momentum together in a single problem.

Key Takeaways

• The strong force holds nucleons together over very short distances and outweighs proton-proton electric repulsion inside the nucleus.
• Every nuclear reaction conserves nucleon number and charge; balance mass numbers (top) and atomic numbers (bottom) on both sides.
• A loss of mass in a reaction releases energy through $E = mc^2$, appearing as kinetic energy of products or as photons like gamma rays.
• Fusion combines light nuclei into a larger one; fission splits a heavy nucleus into smaller ones, and both can release energy.
• Radioactive decay is random for a single nucleus but predictable for a large sample, following $N = N_0 e^{-\lambda t}$.
• Half-life and decay constant are linked by $\lambda = \dfrac{\ln 2}{t_{1/2}}$, and half-life can range from fractions of a second to billions of years.

Physical Properties of Nuclear Interactions

Strong Force in Nucleons

The strong nuclear force is the fundamental interaction that holds nucleons (protons and neutrons) together in atomic nuclei.

• Acts over extremely short distances but is very powerful at nuclear scales
• Overcomes the electromagnetic repulsion between positively charged protons
• Becomes negligible at larger separations, which helps explain why very large nuclei can become unstable
• The strong force controls the interactions of nucleons at nuclear scales

This binding is what makes nuclear reactions possible, since the energy stored in nuclear binding can be released during fission and fusion.

Conservation of Nucleon Number

During any nuclear reaction, the total number of nucleons (protons + neutrons) stays constant.

• In equations, the sum of mass numbers (A) must be equal on both sides
• When a neutron transforms into a proton during beta decay, the total nucleon count stays the same
• This conservation principle lets you balance nuclear equations and predict reaction products
• Example: In the fusion reaction $^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}$, the total nucleon count is 5 on both sides

Conservation Laws in Nuclear Reactions

Nuclear reactions obey the same fundamental conservation laws that govern all physical processes.

• Energy conservation includes both rest-mass energy and kinetic energy of all particles
• Momentum conservation determines the directions and speeds of reaction products
• Charge conservation requires the total electric charge to stay constant

In any nuclear reaction, check the conserved quantities across the whole reaction: nucleon number, charge, total energy, and momentum. Total energy includes both kinetic energy and rest-mass energy, so a decrease in total mass of the products corresponds to released energy according to $E = mc^2$. Because momentum is conserved, emitted particles and daughter nuclei recoil in directions and speeds that keep the total momentum unchanged.

Mass-Energy Equivalence

Einstein's equation $E=mc^2$ is central in nuclear physics, since it explains how energy is released in nuclear reactions.

• The binding energy of a nucleus is the energy equivalent of the mass difference between:
  • The actual nucleus
  • The sum of its separated protons and neutrons
• This "mass defect" ($\Delta m$) multiplied by $c^2$ gives the binding energy
• For all nuclear reactions, mass and energy may be exchanged due to mass-energy equivalence
• In fission and fusion, the products have a slightly lower total mass than the reactants
• This small mass difference converts to a large energy release because $c^2$ is so large

Energy Release in Nuclear Processes

Nuclear reactions release energy in several forms, with the total set by mass-energy equivalence.

• Kinetic energy of reaction products (alpha particles, fission fragments, and so on)
• Electromagnetic radiation (gamma rays)
• The energy released per nucleon is greatest for nuclei with mass numbers around 56 (iron)
• This helps explain why fusion of light elements and fission of heavy elements can both release energy

In nuclear reactions and radioactive decay, released energy may appear as kinetic energy of the products or as photons such as gamma-ray photons.

Nuclear Fusion

Nuclear fusion occurs when two or more smaller nuclei combine to form a larger nucleus and may also produce subatomic particles, releasing energy in the process.

• Requires overcoming the Coulomb barrier (electric repulsion between positively charged nuclei)
• Typically occurs at extremely high temperatures, where particles have enough kinetic energy
• Powers stars like the Sun through hydrogen fusion (application example)
• For example, deuterium-tritium fusion, $^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}$, forms a helium nucleus and a neutron, releasing about 17.6 MeV of energy

Fusion is important in stars and is also studied as a possible energy source (application example).

Nuclear Fission

Nuclear fission is the process by which the nucleus of an atom splits into two or more smaller nuclei, along with subatomic particles, accompanied by energy release.

• Most commonly occurs with heavy elements like uranium and plutonium (application examples)
• The fission products have greater binding energy per nucleon than the original nucleus
• Typically releases a few neutrons per fission event, which can enable chain reactions
• A single uranium-235 fission event releases about 200 MeV of energy (application example)

Spontaneous vs Induced Fission

Whether fission happens spontaneously or needs an energy input depends on the binding-energy situation of the nucleus. If splitting into smaller nuclei lowers the total energy of the system enough, fission can occur spontaneously; otherwise an external input such as neutron absorption is needed to trigger it.

• Spontaneous fission:
  • Occurs naturally in some very heavy, unstable nuclei
  • Happens without external intervention
  • For uranium-238 this process is rare; the commonly cited 4.5 billion year half-life refers to its overall radioactive decay, which is shaped by alpha decay rather than spontaneous fission
• Induced fission:
  • Triggered by the absorption of a particle (usually a neutron)
  • The compound nucleus that forms becomes unstable and splits
  • Example: Uranium-235 readily undergoes induced fission when it captures a neutron (application example)

Radioactive Decay

For a finite sample of radioactive nuclei, the number of undecayed nuclei decreases exponentially with time. This behavior is modeled by $N = N_0 e^{-\lambda t}$, which means equal time intervals remove the same fraction of nuclei, not the same number.

Spontaneous Nuclear Transformation

Radioactive decay is the spontaneous transformation of an unstable nucleus into one or more different nuclei. The exact time at which any single nucleus decays cannot be determined, but for a large sample the decay can be described with probability.

• For a large sample, the decay follows statistical patterns with predictable rates
• The decay of an individual nucleus is random, so only the rate, not the exact moment, can be predicted

Half-life of Radioactive Materials

The half-life describes how quickly a radioactive material decays.

• Defined as the time required for half of the radioactive nuclei in a sample to decay
• Each isotope has a unique half-life; values range from fractions of a second to billions of years
• After one half-life, 50% remains; after two, 25%; after three, 12.5%
• A useful form is $N = N_0 \times 2^{-t/t_{1/2}}$ where:
  • $N$ is the remaining number of radioactive nuclei
  • $N_0$ is the initial number
  • $t$ is the elapsed time
  • $t_{1/2}$ is the half-life

Decay Constant

The decay constant ($\lambda$) represents the probability per unit time that a given nucleus will decay.

• Related to half-life by $\lambda = \frac{\ln 2}{t_{1/2}}$
• Used in the exponential decay equation $N = N_0 e^{-\lambda t}$
• A larger decay constant means a shorter half-life and a faster-decaying sample

A useful rearrangement is $\ln\left(\frac{N}{N_0}\right) = -\lambda t$. This form lets you solve for the age of a sample (when $N_0$ is known) or the time needed for a sample to decay to a given fraction of its original amount.

How to Use This on the AP Physics 2 Exam

Free Response

• Balance nuclear equations by matching the sum of mass numbers (top) and atomic numbers (bottom) on each side, then solve for the unknown.
• When asked for energy released, connect the lost mass to energy with $E = mc^2$ and state where that energy goes (kinetic energy of products or photons).
• For decay problems, choose the form that fits the given information: $N = N_0 e^{-\lambda t}$, $\ln\left(\frac{N}{N_0}\right) = -\lambda t$, or $N = N_0 \times 2^{-t/t_{1/2}}$.
• When you justify a claim, name the conservation law or relationship you are using and explain why the result makes sense.

Problem Solving

• Find the decay constant first when you are given a half-life: $\lambda = \dfrac{\ln 2}{t_{1/2}}$.
• Activity follows the same exponential form as the number of nuclei, so you can use activity values in $A = A_0 e^{-\lambda t}$.
• Keep units consistent. If time is in days, the decay constant comes out in day$^{-1}$.
• Check momentum and charge conservation, not just nucleon number, especially when particles are emitted.

Common Trap

• Do not assume the same number of nuclei decays in each half-life. Each half-life removes half of whatever is left.
• Do not forget that a small mass difference still produces a large energy because $c^2$ is huge.

Common Misconceptions

• "Nuclear reactions create or harm mass and energy out of nothing." Mass and energy are exchanged through $E = mc^2$; the totals are conserved, with a mass decrease showing up as released energy.
• "Each half-life removes the same number of nuclei." Each half-life removes the same fraction (half) of the remaining nuclei, so the actual number removed shrinks over time.
• "You can predict exactly when a single nucleus will decay." Individual decay is random; only the rate for a large sample is predictable.
• "Fusion only happens in labs and fission only happens in reactors." Fusion powers stars and fission can occur spontaneously in some heavy nuclei, so both occur naturally as well as in built systems.
• "The strong force reaches across the whole atom." It acts only over very short, nuclear-scale distances and quickly becomes negligible at larger separations, which is why big nuclei can become unstable.
• "Heavier nuclei always release more energy when they react." The energy released per nucleon peaks near iron-56, which is why light nuclei tend to release energy by fusing and heavy nuclei by splitting.

Practice Problem 1: Conservation in a Nuclear Reaction

Solution

Left side: mass numbers 2 + 3 = 5 and atomic numbers 1 + 1 = 2. Right side: mass numbers 4 + 1 = 5 and atomic numbers 2 + 0 = 2. Therefore, both nucleon number and charge are conserved.

Practice Problem 2: Radioactive Decay

Solution

Use the exponential decay equation: $A = A_0 e^{-\lambda t}$

Where:

• $A$ is the final activity (100 Bq)
• $A_0$ is the initial activity (800 Bq)
• $\lambda$ is the decay constant
• $t$ is the time elapsed (30 days)

(a) To find the decay constant: $100 = 800 e^{-\lambda \times 30}$ $\frac{100}{800} = e^{-30\lambda}$ $0.125 = e^{-30\lambda}$

Taking the natural logarithm of both sides: $\ln(0.125) = -30\lambda$ $-2.0794 = -30\lambda$ $\lambda = \frac{2.0794}{30} = 0.0693 \text{ day}^{-1}$

(b) To find the half-life, use the relationship between decay constant and half-life: $t_{1/2} = \frac{\ln(2)}{\lambda}$ $t_{1/2} = \frac{0.6931}{0.0693} = 10 \text{ days}$

Therefore, the decay constant is 0.0693 day⁻¹ and the half-life is 10 days.

Practice Problem 3: Nuclear Fission

Solution

Apply conservation of nucleon number. The total number of nucleons (protons + neutrons) must be the same on both sides.

Left side:

• Uranium-235: 235 nucleons
• Neutron: 1 nucleon
• Total: 236 nucleons

Right side:

• Krypton-92: 92 nucleons
• Barium-141: 141 nucleons
• x neutrons: x nucleons
• Total: 92 + 141 + x = 233 + x nucleons

Setting these equal: 236 = 233 + x x = 3

Therefore, 3 neutrons are produced in this fission reaction. This matters because these neutrons can go on to cause additional fission events, potentially leading to a chain reaction.

Related AP Physics 2 Guides

• 15.1 Quantum Theory and Wave-Particle Duality
• 15.2 The Bohr Model of Atomic Structure
• 15.3 Emission and Absorption Spectra
• 15.6 Compton Scattering
• 15.8 Types of Radioactive Decay
• 15.4 Blackbody Radiation