20.5 Energy of Nuclear Reactions and Nuclear Binding Energy
PRE-HEALTH PREP
E=mc² in Action
Nuclear Reactions & Nuclear Binding Energy
General Chemistry 20.5
Lesson Overview
Einstein's equation E = Δmc² describes how a small loss of mass during nuclear reactions converts into enormous amounts of energy, as demonstrated through the fission of uranium-235. The mass defect — the difference between the exact masses of reactants and products — provides the key input for calculating nuclear energy release, requiring careful unit conversion from atomic mass units to kilograms. Nuclear binding energy per nucleon serves as the definitive measure of nuclear stability, with iron-56 occupying the peak of the stability curve at 1.38 × 10−¹² J/nucleon.
How E=mc² Governs Nuclear Reaction Energy
Where does all the energy in a nuclear reaction come from? The answer, famously, is mass. Einstein's mass-energy equivalence principle shows that mass and energy are interconvertible, and his equation E = mc² — adapted here as E = Δmc² — quantifies that conversion precisely. You supply the mass change, and the equation tells you exactly how much energy that mass becomes.
The c in the equation is the speed of light: 3.0 × 10⁸ m/s. That is a staggeringly large number, and squaring it makes it even more so. The practical consequence of mass-energy equivalence is that even a tiny amount of mass converts into an enormous quantity of energy — which is exactly why a nuclear power plant can run on what looks like a modest quantity of fuel.
Why so much energy from so little mass? Multiply any mass by (3.0 × 10⁸)² = 9.0 × 10¹⁶ m²/s² and the result is enormous. A single gram of matter, fully converted via E=mc², would release roughly 9 × 10¹³ joules — equivalent to the energy of tens of thousands of tons of conventional explosive.
What Is the Mass Defect in Nuclear Reactions?
The mass defect (Δm) is the difference in mass between the reactants and products of a nuclear reaction. Products always weigh less than the reactants — that "missing" mass is what gets converted into energy according to E=mc².
At first glance, this seems to contradict conservation of mass. If you look at mass numbers — whole-number counts of protons and neutrons — they balance perfectly. Uranium-235 absorbing one neutron gives mass number 236, and the fission products also sum to 236. But mass numbers are rounded integers. When you use exact atomic masses measured to several decimal places, the products are always slightly lighter than the reactants. That small but real discrepancy is the mass defect, and it is the quantity you plug into E = Δmc² to calculate nuclear energy release.
Exact masses vs. mass numbers: Mass numbers (integer counts of nucleons) always balance in a nuclear equation. Exact masses (measured in atomic mass units to many decimal places) do not — and that imbalance is precisely the mass defect you need for nuclear energy calculations.
Worked Example: Energy Released in Uranium-235 Fission
Let's work through a complete calculation for the fission of uranium-235. This nuclear reaction is a textbook example of mass-energy equivalence in action:
Practice Problem: Calculate the amount of energy released for the nuclear fission of one uranium-235 atom.
²³⁵U + ¹n → ¹⁴¹Ba + ⁹¹Kr + 3 ¹n
Step 1: Calculate the Mass Defect
Exact masses (provided):
- Uranium-235: 235.04393 amu
- Neutron: 1.00866 amu
- Barium-141: 141.91643 amu
- Krypton-91: 90.92345 amu
Reactant mass: 235.04393 + 1.00866 = 236.05259 amu
Product mass: 141.91643 + 90.92345 + (3 × 1.00866) = 235.86586 amu
Mass defect: 236.05259 − 235.86586 = 0.18673 amu
Sign convention: Technically, Δm = products − reactants, which gives a negative value. In practice, you take the absolute value and plug it into the energy equation — the result represents energy released.
Step 2: Convert amu to Kilograms
For the energy to come out in joules, Δm must be in kilograms (because a joule is defined as kg·m²/s²). The conversion factor is:
1 amu = 1.67 × 10−²⁷ kg
This factor comes directly from Avogadro's number: since 6.02 × 10²³ amu = 1 gram, one amu equals 1/(6.02 × 10²³) grams = 1.67 × 10−²⁴ g = 1.67 × 10−²⁷ kg. You will normally be provided this conversion on an exam.
Key values for E = Δmc²:
- Δm, the mass defect, must be in kg (1 amu = 1.67 × 10-27 kg)
- c = 3.0 × 108 m/s (speed of light)
Conversion:
0.18673 amu × 1.67 × 10−²⁷ kg/amu = 3.118 × 10−²⁸ kg
Step 3: Apply E = Δmc²
Energy per nucleus:
E = (3.118 × 10−²⁸ kg) × (3.0 × 10⁸ m/s)²
E = 2.81 × 10−¹¹ J per uranium nucleus
Energy per mole of nuclei:
E = 2.81 × 10−¹¹ J × 6.02 × 10²³ = 1.69 × 10¹³ J/mol
The per-nucleus value looks deceptively small. Scale it up to a mole — a mere 235 grams of uranium-235 — and you get 1.69 × 10¹³ joules. That is the energy equivalent of thousands of tons of chemical explosive, released from less than a quarter of a kilogram of material. This is where the power of nuclear reactions becomes viscerally apparent.
Nuclear Binding Energy Explained
The Strong Nuclear Force
The nucleus of an atom is a crowded, positively charged environment. By all rights, the electrostatic repulsion between protons should blow it apart. What holds it together is the strong nuclear force — aptly named because it is far stronger than the electrostatic force it must overcome. The catch is that the strong nuclear force operates only at extremely short distances. Within the tight confines of a nucleus it dominates completely; at longer distances it drops to essentially zero. This dual nature — overwhelming at close range, negligible at larger scales — is why it governs nuclear structure but has no perceptible effect at the atomic or molecular level.
2 Ways the Mass Defect Appears in Nuclear Chemistry
Nuclear binding energy is the energy stored in that strong-force "glue." To calculate it, you use E = Δmc² again — but here the mass defect has a subtly different meaning.
In a nuclear reaction, the mass defect was the difference between the masses of products and reactants. Here, the mass defect for nuclear binding energy is the difference between what a nucleus does weigh and what it should weigh based on its component protons and neutrons. Weigh the nucleus as a whole, then disassemble it and weigh the individual protons and neutrons separately — the separate parts are always heavier. That "missing" mass was converted into the binding energy that holds the nucleus together in the first place.
The mass defect for nuclear binding energy:
Δm = (sum of individual proton and neutron masses) − (measured mass of the intact nucleus)
This difference is always positive: the assembled nucleus is lighter than the sum of its parts.
2 definitions, 1 concept: The mass defect appears in two contexts — nuclear reactions (reactants vs. products) and nuclear binding energy (separated nucleons vs. intact nucleus). In both cases, the "missing" mass has been converted into energy via E = Δmc².
Binding Energy Per Nucleon & Nuclear Stability
Why Total Binding Energy Is Not a Fair Stability Comparison
If you calculate and compare the total nuclear binding energies of iron-56 and uranium-235, uranium's total is larger. But that comparison is meaningless in isolation, because uranium-235 has far more protons and neutrons to hold together than iron-56. You need to normalize for nucleus size.
The standard approach is to express binding energy in joules per nucleon, where a nucleon is any proton or neutron. Divide the total nuclear binding energy by the mass number (the total nucleon count), and you get a fair, size-independent measure of how tightly bound each particle is within its nucleus.
The Babysitter Analogy for Understanding Binding Energy Per Nucleon
Consider two babysitters. You hire the first one at $50/hour to watch all 50 children in your church — that works out to $1/hour per child. You hire the second at $20/hour to watch just your two sons — $10/hour per child. The second babysitter earns far more on a per-child basis, even though the first earns more in total. The per-child rate is the meaningful comparison, not the raw total.
Binding energy per nucleon works the same way. A large nucleus may have a greater total nuclear binding energy while being significantly less stable per nucleon than a smaller one. Per-nucleon binding energy is the metric that matters when assessing nuclear stability.
Why binding energy per nucleon matters physically: A higher binding energy per nucleon means each nucleon is more tightly bound — the nucleus is more stable and less prone to radioactive decay. Iron-56 sits at the global maximum of the nuclear stability curve. Nuclei lighter or heavier than iron are less stable, which is why both fusion (of light nuclei) and fission (of heavy nuclei) can release energy — both reactions produce products closer to iron on the stability curve.
Worked Example: Binding Energy Per Nucleon for Iron-56 vs. Uranium-235
Practice Problem: Calculate the nuclear binding energy per nucleon for iron-56 and uranium-235.
Key values for E = Δmc² (binding energy):
- Δm, the mass defect, must be in kg (1 amu = 1.67 × 10-27 kg)
- c = 3.0 × 108 m/s (speed of light)
| Nuclide / Particle | Mass |
|---|---|
| Proton (1H) | 1.00728 amu |
| Neutron (1n) | 1.00866 amu |
| Uranium-235 (235U) | 235.04393 amu |
| Iron-56 (56Fe) | 55.93494 amu |
Iron-56: The Most Stable Nucleus
Iron-56 contains 26 protons and 30 neutrons. Using exact masses to calculate nuclear binding energy:
- Proton mass: 1.00728 amu
- Neutron mass: 1.00866 amu
Sum of separated nucleon masses:
(26 × 1.00728) + (30 × 1.00866) = 26.18928 + 30.25980 = 56.44908 amu
Measured nuclear mass of iron-56: 55.93494 amu
Mass defect: 56.44908 − 55.93494 = 0.51414 amu
Convert to kg: 0.51414 × 1.67 × 10−²⁷ = 8.586 × 10−²⁸ kg
Total binding energy: (8.586 × 10−²⁸) × (3.0 × 10⁸)² = 7.73 × 10−¹¹ J
Binding energy per nucleon: 7.73 × 10−¹¹ J ÷ 56 = 1.38 × 10−¹² J/nucleon
This value — 1.38 × 10−¹² J/nucleon — is the highest nuclear binding energy per nucleon of any nuclide in existence. Iron-56 is the most tightly bound, and therefore the most stable, nucleus.
Uranium-235: A Less Stable Nucleus
Uranium-235 contains 92 protons and 143 neutrons. Calculating binding energy per nucleon reveals why uranium undergoes fission:
Sum of separated nucleon masses:
(92 × 1.00728) + (143 × 1.00866) = 92.66976 + 144.23838 = 236.90814 amu
Measured nuclear mass of uranium-235: 235.04393 amu
Mass defect: 236.90814 − 235.04393 = 1.86421 amu
Convert to kg: 1.86421 × 1.67 × 10−²⁷ = 3.113 × 10−²⁷ kg
Total binding energy: (3.113 × 10−²⁷) × (3.0 × 10⁸)² = 2.80 × 10−¹⁰ J
Binding energy per nucleon: 2.80 × 10−¹⁰ J ÷ 235 = 1.19 × 10−¹² J/nucleon
Comparing Nuclear Stability: Iron-56 vs. Uranium-235
Side-by-side comparison:
- Iron-56 total binding energy: 7.73 × 10−¹¹ J
- Uranium-235 total binding energy: 2.80 × 10−¹⁰ J (larger total)
- Iron-56 binding energy per nucleon: 1.38 × 10−¹² J/nucleon
- Uranium-235 binding energy per nucleon: 1.19 × 10−¹² J/nucleon (lower per nucleon)
The difference between 1.38 and 1.19 × 10−¹² J/nucleon may look modest, but its physical consequences are dramatic: iron-56 is essentially non-radioactive, while uranium-235 is highly unstable and radioactive. Binding energy per nucleon is your gauge of nuclear stability — the closer a nucleus is to iron's value, the more stable it is; the further below that peak, the more radioactive it is likely to be.
Key insight for nuclear stability: Total nuclear binding energy tells you almost nothing useful about stability — a large, unstable nucleus can have a much larger total binding energy than a small, stable one simply because it contains more nucleons. Always use binding energy per nucleon when assessing nuclear stability.
4-Step Procedure for Calculating Nuclear Binding Energy
These nuclear binding energy calculations are methodical but involve several sequential steps with strict unit requirements. Following a consistent 4-step procedure will help you avoid errors and stay organized on exams.
The 4-Step Method for Binding Energy Per Nucleon
| CALCULATING NUCLEAR BINDING ENERGY (per nucleon) |
|---|
| 1. Calculate the mass defect (Δm = actual mass – predicted mass) |
| 2. Convert to kg (6.02 × 1023 amu = 1g or 1 amu = 1.67 × 10−27 kg) |
| 3. Plug into E = Δmc2 |
| 4. Divide by mass number (mass # = number of nucleons) |
- Calculate the mass defect. Subtract the measured nuclear mass from the sum of the individual proton and neutron masses (take the absolute value).
- Convert to kilograms. Multiply Δm (in amu) by 1.67 × 10−²⁷ kg/amu. This conversion factor will normally be provided.
- Apply E = mc². Multiply the mass in kilograms by the speed of light squared: (3.0 × 10⁸ m/s)². Do not forget to square it.
- Divide by the number of nucleons. Divide the total binding energy by the mass number to get joules per nucleon.
Most common student mistake: Forgetting to square the speed of light in step 3. E = mc² — the exponent applies to c, not the entire product. Always write out (3.0 × 10⁸)² explicitly before calculating.
What your professor is most likely to test: Nuclear binding energy per nucleon — in units of joules per nucleon — is the most common exam question format. Total binding energies may also appear, but per-nucleon values are far more frequently assessed because they carry physical meaning about nuclear stability.
8 Key Takeaways
- In nuclear reactions, a small amount of mass is converted into a large amount of energy according to Einstein's mass-energy equivalence equation E = Δmc², where Δm is the mass defect and c is the speed of light (3.0 × 10⁸ m/s).
- The mass defect in a nuclear reaction is the difference in exact mass between reactants and products. Because mass numbers (integer nucleon counts) balance but exact masses do not, the discrepancy must be measured in amu to several decimal places.
- Before applying E = Δmc², the mass defect must be converted from amu to kilograms using the conversion 1 amu = 1.67 × 10−²⁷ kg.
- For a single nucleus the energy released appears small, but scaling to a mole of nuclei reveals enormous energy output — illustrating why nuclear fuel is so extraordinarily energy-dense.
- Nuclear binding energy is the energy associated with the strong nuclear force that holds a nucleus together. Its mass defect is the difference between the summed masses of the separated protons and neutrons and the measured mass of the intact nucleus.
- Binding energy per nucleon (total nuclear binding energy ÷ mass number) is the correct metric for comparing nuclear stability. A higher value indicates a more stable, less radioactive nucleus.
- Iron-56 has the highest binding energy per nucleon of any nucleus (1.38 × 10−¹² J/nucleon) and is the most stable nuclide. Uranium-235 (1.19 × 10−¹² J/nucleon) is far less stable and highly radioactive.
- Never forget to square the speed of light in step 3 of the calculation — this is the most common arithmetic error in nuclear binding energy problems.