Nuclear Energy Resources: Mass-Energy, Binding Energy, Radioactive Decay, Fission and Fusion

This chapter, from the Engineering Sciences (XE) paper's Energy Science material, answers the second topic of Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam: “Nuclear energy resources: Energy-mass relation, nuclear reaction conservation laws, binding energy and Q-values, radioactive decay, fission and fusion.” It teaches the mass-energy relation, the conservation laws, binding energy and the binding-energy curve, Q-values of reactions and decays, decay law, activity and decay chains in outline, the fission chain reaction with moderators and criticality in outline, and fusion. Atomic masses are always given in the question stems. The numericals are mass defect, binding energy per nucleon, Q-value, alpha energy, activity, equilibrium ratio, fuel consumption of a reactor and mass converted.

1. Mass-energy relation and conservation laws

Einstein's relation E = mc² says that a mass m is equivalent to an energy mc². A mass change Δm in a nuclear process appears as an energy change ΔE = Δm c². The convenient unit is the unified atomic mass unit: 1 u = 931.5 MeV/c² (u is one-twelfth of the mass of a ¹²C atom), with 1 MeV = 1.602 × 10⁻¹³ J. Nuclear energies are millions of times chemical energies because the energy per nucleon exchanged is of order MeV, against eV per atom for chemical bonds.

Worked example: a power plant releasing 1000 MW of heat for one year (3.1536 × 10⁷ s) releases 3.1536 × 10¹⁶ J. Taking c = 3.00 × 10⁸ m/s, the mass that disappears is Δm = E/c² = 3.1536 × 10¹⁶/9.0 × 10¹⁶ = 0.35 kg. Only a fraction of a kilogram of the fuel mass is converted, although tonnes of material are handled.

In every nuclear reaction these are conserved: electric charge (sum of Z), baryon number (sum of mass numbers A, that is the number of nucleons), total energy including rest-mass energy, linear momentum, angular momentum and lepton number. The total number of protons and the total number of neutrons are separately conserved in reactions driven by the strong interaction (fission, deuterium-tritium fusion, neutron capture), but not in beta decay or in any weak-interaction step such as the first step of the solar chain, where a neutron turns into a proton (or the reverse) while A stays the same. Rest mass and kinetic energy are not separately conserved; only their sum is.

🧠 Memory trick
To complete a reaction, balance the top numbers (A) and the bottom numbers (Z) separately: ¹⁴₇N + ⁴₂He → ¹⁷₈O + X needs A = 18 − 17 = 1 and Z = 9 − 8 = 1, so X is a proton ¹₁H.

2. Binding energy and the binding-energy curve

A nucleus of Z protons and N neutrons weighs less than its separate nucleons. The mass defect is Δm = Z m_p + N m_n − M_nucleus and the binding energy is B = Δm c², the energy needed to take the nucleus apart. In practice use atomic masses: B = [Z M(¹H) + N m_n − M(atom)] c², because the Z electrons of the hydrogen atoms cancel the Z electrons of the atom (their small binding energies are neglected). With M(¹H) = 1.007825 u and m_n = 1.008665 u.

Worked example for ⁴He (atomic mass 4.002603 u): Δm = 2(1.007825) + 2(1.008665) − 4.002603 = 0.030377 u, so B = 0.030377 × 931.5 = 28.30 MeV and B/A = 7.07 MeV per nucleon. For ⁵⁶Fe (55.934936 u, Z = 26, N = 30): Δm = 0.528464 u, B = 492.3 MeV and B/A = 8.79 MeV per nucleon.

Region of the curveB/A behaviourConsequence
Light nuclei (A below about 20)Rises steeply, with peaks at ⁴He and similar nucleiFusion releases energy
Near A of 56 to 62 (iron group)Maximum, about 8.8 MeV per nucleonMost tightly bound; no net gain from fusing or splitting
Heavy nuclei (A above about 100)Falls slowly, to about 7.6 MeV per nucleon at uraniumFission releases energy
🎯 Why it matters
B/A falls at high A because Coulomb repulsion grows as Z² while the short-range nuclear attraction grows only with A; at low A the surface nucleons are poorly bound. The balance puts the peak in the iron group. Splitting ²³⁵U (about 7.6 MeV/nucleon) into fragments near A of 120 (about 8.5 MeV/nucleon) gains about 0.9 × 235, which is roughly 200 MeV.

3. Q-values

For a reaction a + X → Y + b the Q-value is the energy released: Q = [Σ m(reactants) − Σ m(products)] c² = (kinetic energy of products) − (kinetic energy of reactants). Q > 0 is exoergic (releases energy, mass decreases); Q < 0 is endoergic and needs at least a threshold kinetic energy, always more than |Q| because some energy stays in the centre-of-mass motion. Use atomic masses whenever the numbers of electrons balance on the two sides.

Worked example, the deuterium-tritium reaction ²H + ³H → ⁴He + n with atomic masses 2.014102, 3.016049, 4.002603 and 1.008665 u: reactants 5.030151 u, products 5.011268 u, Δm = 0.018883 u, Q = 0.018883 × 931.5 = 17.59 MeV. Momentum conservation (reactants nearly at rest) shares this inversely with the mass: the neutron (mass 1) takes about 4/5, 14.1 MeV, and the alpha (mass 4) about 3.5 MeV.

DecayQ in atomic massesNote
Alpha[M(P) − M(D) − M(⁴He)] c²Alpha carries Q × M_D/(M_D + M_α), about Q × (A − 4)/A
Beta minus[M(P) − M(D)] c²Shared by the electron and antineutrino
Beta plus[M(P) − M(D) − 2 m_e] c², 2m_e c² = 1.022 MeVNeeds the parent to be heavier by more than 2 m_e
⚠️ Exam trap
In beta-minus decay with atomic masses, no electron mass is subtracted; in beta-plus decay 2m_e is subtracted. Using the wrong rule gives a Q off by 0.511 or 1.022 MeV. Example: ¹⁴C (14.003242 u) → ¹⁴N (14.003074 u) has Q = 0.000168 × 931.5 = 0.157 MeV.

4. Radioactive decay, activity and decay chains

Decay is random, each nucleus having the same probability λ per unit time (the decay constant): dN/dt = −λN, so N = N₀ e^(−λt). The activity A = λN = A₀ e^(−λt) is the number of decays per second, in becquerel (Bq = 1 s⁻¹); 1 curie = 3.7 × 10¹⁰ Bq. The half-life is T½ = ln 2/λ = 0.693/λ and the mean life τ = 1/λ = 1.443 T½. After n half-lives the fraction left is 2⁻ⁿ; in general t = T½ log₂(N₀/N).

Worked example: 1 g of ²²⁶Ra (T½ = 1600 years, 1 year = 3.156 × 10⁷ s, molar mass 226 g/mol, N_A = 6.022 × 10²³). λ = 0.6931/(1600 × 3.156 × 10⁷) = 1.373 × 10⁻¹¹ s⁻¹; N = 6.022 × 10²³/226 = 2.665 × 10²¹; A = λN = 3.66 × 10¹⁰ Bq, about one curie. Dating example: if 30 per cent of the ¹⁴C (T½ = 5730 years) remains, t = 5730 × log₂(1/0.30) = 9950 years.

Decay chains. A parent P decays to a daughter D that is itself radioactive. The daughter's activity builds up and then follows the parent. If the parent is longer-lived than the daughter (λ_P < λ_D) and enough time has passed, transient equilibrium sets in with A_D/A_P = λ_D/(λ_D − λ_P), which is greater than 1. If the parent is very much longer-lived (λ_P ≪ λ_D), secular equilibrium gives A_D ≈ A_P. If the parent is the shorter-lived one, no equilibrium is reached. Worked example: T½(P) = 10 h and T½(D) = 2 h; since λ is proportional to 1/T½, A_D/A_P = (1/2)/(1/2 − 1/10) = 1.25. The natural heavy-element chains end in stable isotopes of lead.

5. Fission and the chain reaction

In fission a heavy nucleus such as ²³⁵U, after absorbing a neutron, splits into two fragments with the release of about 2.4 neutrons on average (for ²³⁵U with thermal neutrons) and about 200 MeV per fission. Most of this energy is the kinetic energy of the fragments, which becomes heat. In terms of mass, 200 MeV × 1.602 × 10⁻¹³ J/MeV × 6.022 × 10²³/235 g gives about 8.2 × 10¹⁰ J per gram, or 82 TJ per kg of ²³⁵U fully fissioned, a few million times the heat of burning coal (order of tens of MJ/kg).

Worked example: a reactor of 3000 MW thermal power, 200 MeV per fission, 3.204 × 10⁻¹¹ J per fission. Fissions per second = 3 × 10⁹/3.204 × 10⁻¹¹ = 9.36 × 10¹⁹. Per day (86 400 s) = 8.09 × 10²⁴ fissions = 13.4 mol = 3157 g, so about 3.16 kg of ²³⁵U is fissioned per day.

The multiplication factor k is the ratio of neutrons in one generation to the number in the previous one. k < 1 is subcritical, k = 1 critical (steady power) and k > 1 supercritical; the reactivity is ρ = (k − 1)/k. After n generations the population is N₀kⁿ, so with a generation time τ_g it is N₀ k^(t/τ_g). Worked example: k = 1.002 and τ_g = 0.1 s give 1.002¹⁰⁰ = 1.22 after 10 s. Criticality is reached only if production beats losses, which depends on the fuel enrichment, the size and shape (leakage falls as the surface-to-volume ratio falls), a reflector and the absorbers present.

Fission neutrons are fast (about 2 MeV) but ²³⁵U fissions far more readily with thermal neutrons (about 0.025 eV). A moderator slows neutrons by elastic collisions; it should have light nuclei (a collision with a light nucleus removes more energy) and low neutron absorption. Ordinary water, heavy water and graphite are used. Control rods (strong absorbers such as boron or cadmium) lower k when inserted. The small fraction of delayed neutrons (under one per cent of the total) stretches the effective generation time so that control by rods is practicable. Natural uranium is about 0.7 per cent ²³⁵U; ²³⁸U and ²³²Th are fertile (they breed fissile ²³⁹Pu and ²³³U on neutron capture).

6. Fusion

In fusion light nuclei join and release energy because B/A rises toward the iron group. The reaction easiest to achieve is deuterium-tritium, with Q = 17.6 MeV. Per unit mass fusion fuel gives more than fission fuel: using mass numbers as masses, D-T gives 17.6/5 = 3.52 MeV per nucleon of fuel against 200/235 = 0.851 for ²³⁵U fission, a ratio of 4.1. The solar fusion chain turns four ¹H into one ⁴He, releasing about 26.7 MeV in all.

The obstacle is the Coulomb barrier: two positive nuclei repel, with a barrier of the order of an MeV or more. Fusion therefore needs temperatures of order 10⁸ K (tens of keV), where the fuel is a plasma, helped by quantum tunnelling and the fast tail of the energy distribution. A power-producing plasma must satisfy the Lawson criterion: the product of density and confinement time (and, in the triple-product form, of temperature too) must exceed a threshold so that fusion heating beats losses. Confinement is by magnetic fields (tokamak) or by compressing a pellet very quickly (inertial). Deuterium is abundant in water; tritium is scarce and is bred from lithium by neutron capture, ⁶Li + n → ⁴He + ³H.

Key takeaways

  • E = mc² with 1 u = 931.5 MeV/c². Charge, nucleon number, total energy, momentum and angular momentum are conserved; rest mass and kinetic energy separately are not.
  • B = [Z M(¹H) + N m_n − M(atom)] c²; B/A peaks near iron (about 8.8 MeV per nucleon), so light nuclei fuse and heavy nuclei fission with release of energy.
  • Q = (Σ m reactants − Σ m products) c²; Q > 0 is exoergic. Beta-plus decay subtracts 2m_e (1.022 MeV); beta-minus does not. Alpha energy is about Q × (A − 4)/A.
  • N = N₀e^(−λt), A = λN, T½ = 0.693/λ. In a chain, transient equilibrium gives A_D/A_P = λ_D/(λ_D − λ_P); secular equilibrium gives A_D ≈ A_P.
  • Fission gives about 200 MeV and about 2.4 neutrons per ²³⁵U fission. k < 1, = 1, > 1 means subcritical, critical, supercritical; moderators slow neutrons, control rods absorb them, and delayed neutrons make control possible.
  • D-T fusion gives 17.6 MeV (14.1 to the neutron); it is hindered by the Coulomb barrier and needs about 10⁸ K plasma satisfying the Lawson criterion; per unit mass it beats fission about fourfold.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following nuclear reactions is allowed, as far as charge and mass number (nucleon number) are concerned?

    1. ¹⁴₇N + ⁴₂He → ¹⁶₈O + ¹₁H, since the 18 nucleons on the left are said to give 17 nucleons on the right and charge is preserved
    2. ²³⁸₉₂U → ²³⁴₉₁Pa + ⁴₂He, which keeps the mass number balanced but changes the total charge on the two sides
    3. ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He, in which the thorium and helium numbers add to the uranium numbers
    4. ²H + ³H → ⁴He + 2n, which is a fusion equation that balances the charge but not the mass number
    Show answer

    Answer: C — ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He, in which the thorium and helium numbers add to the uranium numbers

    For uranium to thorium plus helium, A: 238 = 234 + 4 and Z: 92 = 90 + 2. The protactinium case has 91 + 2 = 93 ≠ 92; the nitrogen-oxygen case has 16 + 1 = 17 ≠ 18; and the D-T case has A: 5 → 6 and Z: 2 → 2, so A fails.

  2. Why can both the fusion of very light nuclei and the fission of very heavy nuclei release energy?

    1. Binding energy per nucleon is largest for the lightest and the heaviest nuclei, so energy is always gained by moving away from the middle of the periodic table
    2. Both processes convert neutrons into protons in the products, and that always lowers the total mass of the system
    3. Binding energy per nucleon peaks in the iron region, so moving toward that region from either side increases binding and releases energy
    4. Both processes increase the Coulomb repulsion between the products, which is then stored as extra kinetic energy
    Show answer

    Answer: C — Binding energy per nucleon peaks in the iron region, so moving toward that region from either side increases binding and releases energy

    The B/A curve rises to a maximum near A of 56 to 62. Fusion of light nuclei and fission of heavy nuclei both move products toward higher B/A and so release energy. The curve does not peak at the two ends.

  3. The atomic mass of deuterium ²H is 2.014102 u, that of ¹H is 1.007825 u and that of the neutron is 1.008665 u; take 1 u = 931.5 MeV/c². The binding energy per nucleon of ²H is closest to:

    1. 2.22 MeV
    2. 1.11 MeV
    3. 7.07 MeV
    4. 0.56 MeV
    Show answer

    Answer: B — 1.11 MeV

    Δm = 1.007825 + 1.008665 − 2.014102 = 0.002388 u, so B = 0.002388 × 931.5 = 2.22 MeV; per nucleon (A = 2) it is 1.11 MeV. The value 2.22 MeV is the total, and 7.07 MeV belongs to ⁴He.

  4. For the reaction ²H + ³H → ⁴He + n the atomic masses are 2.014102 u, 3.016049 u, 4.002603 u and 1.008665 u, and 1 u = 931.5 MeV/c². The Q-value is closest to:

    1. +26.7 MeV, released
    2. −17.6 MeV, absorbed
    3. +5.0 MeV, released
    4. +17.6 MeV, released
    Show answer

    Answer: D — +17.6 MeV, released

    Reactants 5.030151 u minus products 5.011268 u = 0.018883 u, times 931.5 = +17.6 MeV; mass decreases, so energy is released. The −17.6 value has the sign wrong; 26.7 MeV belongs to the hydrogen-to-helium chain.

  5. ²²⁶Ra decays by alpha emission to ²²²Rn with Q = 4.87 MeV. Treating the parent as at rest and using non-relativistic momentum conservation with mass numbers 222 for the daughter and 4 for the alpha, the kinetic energy of the alpha particle is closest to:

    1. 4.87 MeV, all of Q
    2. 2.44 MeV, half of Q
    3. 0.09 MeV
    4. 4.78 MeV
    Show answer

    Answer: D — 4.78 MeV

    Equal and opposite momenta make kinetic energy inversely proportional to mass, so the alpha takes Q × 222/226 = 4.78 MeV and the heavy daughter recoils with about 0.09 MeV. Giving the whole Q to the alpha ignores recoil.

  6. The activity of a pure radioactive sample falls from 800 Bq to 100 Bq in 30 days. Its half-life is:

    1. 3.75 days
    2. 15 days
    3. 10 days
    4. 7.5 days
    Show answer

    Answer: C — 10 days

    800 to 100 is a factor of 8 = 2³, so three half-lives take 30 days and T½ = 10 days. Dividing 30 days by the factor 8 is the slip behind 3.75.

  7. A long-lived parent (half-life thousands of years) decays to a short-lived daughter (half-life of days), and the sample has been left undisturbed for many daughter half-lives. Which statement holds?

    1. The daughter activity is almost equal to the parent activity (secular equilibrium)
    2. The daughter has decayed away completely, because its half-life is shorter
    3. The number of daughter atoms equals the number of parent atoms
    4. The daughter activity is far greater than the parent activity because λ_D is larger
    Show answer

    Answer: A — The daughter activity is almost equal to the parent activity (secular equilibrium)

    The daughter is replenished as fast as it decays, so λ_D N_D ≈ λ_P N_P and the activities match (the numbers do not, since N_D/N_P = λ_P/λ_D is tiny). The daughter does not vanish while the parent supplies it.

  8. Why is a moderator in a thermal reactor made of light nuclei with a low neutron-absorption cross-section?

    1. Light nuclei are chosen only because they are cheap, since the energy transferred in a collision is independent of nuclear mass
    2. A heavy nucleus would capture every neutron that reaches it, and a light nucleus always releases additional fission neutrons of its own
    3. Light nuclei make the neutrons move faster after each collision, which in turn raises the fission cross-section of ²³⁵U
    4. A light nucleus takes up more of the neutron's energy per elastic collision, and low absorption avoids wasting the neutrons that are being slowed
    Show answer

    Answer: D — A light nucleus takes up more of the neutron's energy per elastic collision, and low absorption avoids wasting the neutrons that are being slowed

    A neutron colliding with a nucleus of similar mass can lose a large fraction of its energy, while off a heavy nucleus it bounces with little loss. Slow neutrons have the larger ²³⁵U fission cross-section, so the moderator must slow them without absorbing them.

  9. A reactor has multiplication factor k = 1.002 and an effective generation time of 0.1 s. Starting from a steady neutron population N₀, the population after 10 s is closest to:

    1. 1.02 N₀
    2. 1.22 N₀
    3. 7.4 N₀
    4. 2.00 N₀
    Show answer

    Answer: B — 1.22 N₀

    10 s is 100 generations, so N = N₀ × 1.002¹⁰⁰ = N₀ e^(100 × 0.001998) = 1.22 N₀. Multiplying 1.002 by 10 s gives the wrong 1.02; compounding is essential.

  10. Why do fusion reactions need temperatures of the order of 10⁸ K although the nuclei are ordinarily bound only by a few MeV?

    1. The nuclei must overcome their mutual Coulomb repulsion, which needs large relative kinetic energy
    2. The nuclei must first be heated enough to break up their own binding energy before they can join together
    3. Free neutrons are needed in the plasma, and they are produced only at very high temperature
    4. The reaction has a negative Q-value, and the supplied heat provides the missing mass
    Show answer

    Answer: A — The nuclei must overcome their mutual Coulomb repulsion, which needs large relative kinetic energy

    Two nuclei carry positive charge and repel; they must approach to nuclear range for the strong force to act, which requires high relative kinetic energy, hence a hot plasma. D-T fusion has a large positive Q.

  11. Which quantities are conserved in every nuclear reaction?

    1. Total electric charge
    2. Total number of nucleons (mass number)
    3. Total rest mass of the particles
    4. Total kinetic energy of the particles
    Show answer

    Answer: A — Total electric charge; B — Total number of nucleons (mass number)

    Charge and nucleon number are conserved, along with total energy and momentum. Rest mass and kinetic energy are exchanged in any reaction with a non-zero Q, so neither is conserved separately.

  12. Which statements about the binding-energy-per-nucleon curve are correct?

    1. It has a broad maximum near mass numbers 56 to 62
    2. The fission fragments of ²³⁵U have higher B/A than ²³⁵U itself
    3. B/A increases steadily with mass number all the way to uranium
    4. ⁴He has a B/A much lower than its neighbouring light nuclei such as ²H
    Show answer

    Answer: A — It has a broad maximum near mass numbers 56 to 62; B — The fission fragments of ²³⁵U have higher B/A than ²³⁵U itself

    The maximum is in the iron group, and fragments near A of 120 are more tightly bound per nucleon than uranium, which is why fission releases energy. B/A falls beyond the maximum, and ⁴He (7.07) is far above ²H (1.11).

  13. Which statements about fission chain reactions are correct?

    1. A reactor operating at steady power has an effective multiplication factor k = 1
    2. Delayed neutrons, a small fraction of the total, lengthen the effective generation time and make control practicable
    3. The moderator speeds up fission neutrons so that they reach ²³⁵U sooner
    4. Inserting boron or cadmium control rods raises k by adding neutron-producing nuclei
    Show answer

    Answer: A — A reactor operating at steady power has an effective multiplication factor k = 1; B — Delayed neutrons, a small fraction of the total, lengthen the effective generation time and make control practicable

    Steady power means k = 1, and delayed neutrons stretch the response time. A moderator slows neutrons and rods absorb them, lowering k, so the last two statements are false.

  14. Using atomic masses M(⁴He) = 4.002603 u, M(¹H) = 1.007825 u and m_n = 1.008665 u, and 1 u = 931.5 MeV/c², find the binding energy per nucleon of ⁴He in MeV. Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 7.07

    Δm = 2(1.007825) + 2(1.008665) − 4.002603 = 0.030377 u. B = 0.030377 × 931.5 = 28.296 MeV. B/A = 28.296/4 = 7.074, which is 7.07 MeV. Python check gives 7.0741.

  15. The half-life of ²²⁶Ra is 1600 years. Take 1 year = 3.156 × 10⁷ s, molar mass 226 g/mol and N_A = 6.022 × 10²³ mol⁻¹. Find the activity of 1.00 g of ²²⁶Ra in units of 10¹⁰ Bq, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 3.66

    λ = 0.693147/(1600 × 3.156 × 10⁷) = 1.3727 × 10⁻¹¹ s⁻¹. N = 6.022 × 10²³/226 = 2.6646 × 10²¹. A = λN = 3.658 × 10¹⁰ Bq, which is 3.66 in units of 10¹⁰ Bq (Python gives 3.6576).

  16. A reactor produces 3000 MW of thermal power entirely from fission of ²³⁵U, with 200 MeV released per fission. Take 1 MeV = 1.602 × 10⁻¹³ J, N_A = 6.022 × 10²³ mol⁻¹, a molar mass of 235 g/mol and 86 400 s per day, and ignore neutron capture without fission. How many kilograms of ²³⁵U are fissioned per day? Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 3.16

    Energy per fission = 200 × 1.602 × 10⁻¹³ = 3.204 × 10⁻¹¹ J. Fissions per second = 3 × 10⁹/3.204 × 10⁻¹¹ = 9.363 × 10¹⁹. Per day: 8.090 × 10²⁴, which is 13.43 mol and 3157 g = 3.16 kg. Python gives 3.157.

  17. A 1000 MW (thermal) power plant runs for one year of 3.1536 × 10⁷ s. Take c = 3.00 × 10⁸ m/s. Find the mass, in kilograms, that is converted into energy. Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.35

    E = 10⁹ × 3.1536 × 10⁷ = 3.1536 × 10¹⁶ J. Δm = E/c² = 3.1536 × 10¹⁶/9.00 × 10¹⁶ = 0.3504 kg, which is 0.35 kg.

  18. A parent isotope of half-life 10 hours decays to a daughter of half-life 2 hours. After a long time the system is in transient equilibrium. Find the ratio of the daughter activity to the parent activity, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 1.25

    In transient equilibrium A_D/A_P = λ_D/(λ_D − λ_P). With λ proportional to 1/T½ (the factor ln 2 cancels): (1/2)/(1/2 − 1/10) = 0.5/0.4 = 1.25. Python check gives 1.25.