Rutherford Alpha Scattering, Exponential Decay N = N₀e⁻ᵏᵗ, Mass Defect & Binding Energy per Nucleon
Nuclear physics explores the structure, stability, and energy of atomic nuclei, beginning with Rutherford's alpha-scattering experiment which proved that an atom is mostly empty space with a tiny, dense, positively charged nucleus at its centre. High-energy electron diffraction reveals that nuclear radius obeys R = R₀ A^(1/3) (where R₀ ≈ 1.05-1.20 fm), proving that all nuclei have a constant, immense nuclear density (~10¹⁷ kg m⁻³). Unstable nuclei decay randomly and spontaneously toward the N-Z stability belt according to Activity A = -ΔN / Δt = λN and N = N₀ e^(-λt). Finally, Einstein's mass-energy equivalence (ΔE = Δmc², where 1 u = 931.5 MeV) explains why the Binding Energy per Nucleon curve peaks at Iron-56 (~8.8 MeV), allowing energy to be released both by nuclear fusion of light nuclei and by induced nuclear fission of heavy Uranium-235.
- Rutherford Alpha Scattering, Closest Approach & Electron Diffraction
- Firing monoenergetic α-particles at a thin gold foil (~1 atom / few hundred atoms thick, in a vacuum so α-particles aren't absorbed by air): 1. Most (>99%) pass straight through undeviated → atom is mostly empty space. 2. A small fraction (~1 in 2000) deflect through small angles → nucleus is positively charged (electrostatic repulsion). 3. A tiny fraction (~1 in 8000) rebound backward (>90°) → almost all the atom's mass and positive charge are concentrated in a tiny, dense nucleus. Distance of closest approach (r_min, an upper limit on nuclear radius): in a head-on collision, initial kinetic energy E_k converts completely into electric potential energy: E_k = (Q_α Q_nucleus) / (4πε₀ r_min). High-energy electron diffraction (λ = h / p ≈ 1 fm) is superior to α-particles for measuring nuclear radius because electrons do not experience the strong nuclear force and give a clear first diffraction minimum at sin θ ≈ 0.61λ / R.
- Nuclear Radius (R = R₀ A^(1/3)) & Constant Nuclear Density
- Nuclear radius R is directly proportional to the cube root of nucleon number A: R = R₀ A^(1/3), where R₀ ≈ 1.05 × 10⁻¹⁵ m (1.05 fm). Proof that nuclear density ρ is independent of nucleon number A: Assuming a spherical nucleus of volume V = (4/3)πR³ = (4/3)π(R₀ A^(1/3))³ = (4/3)πR₀³ A, and mass M = A u (where u = 1.661 × 10⁻²⁷ kg), Density ρ = Mass / Volume = (A u) ÷ [(4/3)πR₀³ A] = 3u / (4πR₀³) ≈ 2.3 × 10¹⁷ kg m⁻³. Notice that nucleon number A cancels out completely, proving nuclear matter has constant density and nucleons are incompressible.
- Alpha, Beta & Gamma Radiation, Inverse-Square Law & N-Z Stability Graph
- 1. Alpha (⁴₂α): 2p + 2n, strongly ionising, stopped by paper or ~5 cm of air. 2. Beta-minus (⁰₋₁β⁻): fast electron from neutron-rich nuclei (above N-Z stability line), stopped by ~3 mm of aluminium. 3. Beta-plus (⁰₊₁β⁺): positron from proton-rich nuclei (below N-Z stability line), annihilates with electrons. 4. Gamma (γ): high-energy EM photon emitted when an excited daughter nucleus de-excites from a metastable energy level; weakly ionising, reduced by thick lead and obeys the inverse-square law in air: I = k / (x + x₀)² (subtract background count rate C_b first. Plotting 1/√C against x gives a straight line). Light stable nuclei have N = Z; heavy stable nuclei (Z > 20) curve above N = Z (up to N ≈ 1.5 Z) because more neutrons are needed to add strong nuclear attraction without adding electrostatic proton repulsion.
Rutherford Alpha Scattering, Exponential Decay N = N₀e⁻ᵏᵗ, Mass Defect & Binding Energy per Nucleon: Exponential Half-Life Curve & Radiation Shielding
Adjust the number of elapsed half-lives to calculate undecayed nuclei N = N₀ × (½)ⁿ and compare alpha, beta, and gamma absorption.
Key Concepts & Exam Definitions (6)
- Rutherford Alpha Scattering, Closest Approach & Electron Diffraction
- Firing monoenergetic α-particles at a thin gold foil (~1 atom / few hundred atoms thick, in a vacuum so α-particles aren't absorbed by air): 1. Most (>99%) pass straight through undeviated → atom is mostly empty space. 2. A small fraction (~1 in 2000) deflect through small angles → nucleus is positively charged (electrostatic repulsion). 3. A tiny fraction (~1 in 8000) rebound backward (>90°) → almost all the atom's mass and positive charge are concentrated in a tiny, dense nucleus. Distance of closest approach (r_min, an upper limit on nuclear radius): in a head-on collision, initial kinetic energy E_k converts completely into electric potential energy: E_k = (Q_α Q_nucleus) / (4πε₀ r_min). High-energy electron diffraction (λ = h / p ≈ 1 fm) is superior to α-particles for measuring nuclear radius because electrons do not experience the strong nuclear force and give a clear first diffraction minimum at sin θ ≈ 0.61λ / R.
- Nuclear Radius (R = R₀ A^(1/3)) & Constant Nuclear Density
- Nuclear radius R is directly proportional to the cube root of nucleon number A: R = R₀ A^(1/3), where R₀ ≈ 1.05 × 10⁻¹⁵ m (1.05 fm). Proof that nuclear density ρ is independent of nucleon number A: Assuming a spherical nucleus of volume V = (4/3)πR³ = (4/3)π(R₀ A^(1/3))³ = (4/3)πR₀³ A, and mass M = A u (where u = 1.661 × 10⁻²⁷ kg), Density ρ = Mass / Volume = (A u) ÷ [(4/3)πR₀³ A] = 3u / (4πR₀³) ≈ 2.3 × 10¹⁷ kg m⁻³. Notice that nucleon number A cancels out completely, proving nuclear matter has constant density and nucleons are incompressible.
- Alpha, Beta & Gamma Radiation, Inverse-Square Law & N-Z Stability Graph
- 1. Alpha (⁴₂α): 2p + 2n, strongly ionising, stopped by paper or ~5 cm of air. 2. Beta-minus (⁰₋₁β⁻): fast electron from neutron-rich nuclei (above N-Z stability line), stopped by ~3 mm of aluminium. 3. Beta-plus (⁰₊₁β⁺): positron from proton-rich nuclei (below N-Z stability line), annihilates with electrons. 4. Gamma (γ): high-energy EM photon emitted when an excited daughter nucleus de-excites from a metastable energy level; weakly ionising, reduced by thick lead and obeys the inverse-square law in air: I = k / (x + x₀)² (subtract background count rate C_b first. Plotting 1/√C against x gives a straight line). Light stable nuclei have N = Z; heavy stable nuclei (Z > 20) curve above N = Z (up to N ≈ 1.5 Z) because more neutrons are needed to add strong nuclear attraction without adding electrostatic proton repulsion.
- Radioactive Decay Law (A = λN), Exponential Decay & Half-Life (T₁/₂ = ln(2) / λ)
- Radioactive decay is spontaneous (unaffected by temperature/pressure/chemical bonding) and random (impossible to predict when a specific nucleus will decay; each nucleus has a constant probability of decay per unit time, the decay constant λ in s⁻¹). Activity A (decays per second, in Becquerels Bq = s⁻¹) is proportional to the number of undecayed nuclei N remaining: A = -ΔN / Δt = λN (where N = (m / M_r) × N_A). Exponential equations: N = N₀ e^(-λt), A = A₀ e^(-λt), and Corrected Count Rate C = C₀ e^(-λt). Half-life T₁/₂ is the average time taken for the number of undecayed nuclei (or activity) to halve: T₁/₂ = ln(2) / λ ≈ 0.693 / λ.
- Mass Defect (Δm), Binding Energy & The Binding Energy per Nucleon Curve
- The mass of any nucleus is always less than the sum of the masses of its individual separated protons and neutrons. This difference is the Mass Defect: Δm = [Z m_p + (A - Z) m_n] - m_nucleus. Binding Energy (BE) is the work that must be done (energy supplied) to completely separate a nucleus into its constituent protons and neutrons (or the energy released when the nucleus is formed from its nucleons): BE = Δm c² (in Joules, when Δm is in kg) OR BE (in MeV) = Δm (in atomic mass units u) × 931.5 MeV u⁻¹. The graph of Binding Energy per Nucleon (BE / A) against nucleon number A rises steeply for light nuclei (with a sharp stability spike at ⁴₂He) to a maximum peak of ~8.8 MeV at ⁵⁶₂₆Fe (the most stable nucleus), then decreases gradually toward ~7.6 MeV at ²³⁸₉₂U. In any nuclear reaction (fission, fusion, or α/β decay), energy released Q = Σ(BE of products) - Σ(BE of reactants) = [Σ(mass of reactants) - Σ(mass of products)] × c².
- Induced Nuclear Fission (Thermal Reactors) vs Nuclear Fusion
- Both fission of heavy nuclei (A > 56) and fusion of light nuclei (A < 56) move nuclei UP the curve toward the Iron-56 peak, increasing binding energy per nucleon and therefore releasing energy. 1. Induced Fission in a Pressurised Water Reactor (PWR): a slow-moving ('thermal') neutron (E_k ~0.03 eV) is absorbed by a ²³⁵₉₂U nucleus, forming unstable ²³⁶₉₂U which splits into two neutron-rich daughter fragments + 2 or 3 fast neutrons + ~200 MeV kinetic energy. Moderator (water or graphite): slows fast neutrons down via elastic collisions (most kinetic energy transferred when moderator nuclei have small mass close to a neutron). Control rods (boron or cadmium): absorb excess neutrons so exactly 1 neutron per fission triggers another fission (criticality; dropped fully in an emergency shutdown). Coolant (high-pressure water): transfers heat to steam turbines. 2. Nuclear Fusion: two light positively charged nuclei (e.g. ²₁H + ³₁H → ⁴₂He + ¹₀n) fuse if heated to extreme plasma temperatures (~10⁸ K, high E_k) and high density so they overcome electrostatic Coulomb repulsion and come within ~1-3 fm where the strong nuclear force attracts them.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State two reasons why high-energy electron diffraction gives a much more accurate measurement of nuclear radius than alpha-particle closest approach.
Show mathematically why all atomic nuclei have the same density regardless of their nucleon number A.
Why does the N-Z stability curve bend upward above the N = Z line for heavy nuclei (Z > 20)?
Define the mass defect and binding energy of a nucleus, and state how to convert mass defect in atomic mass units (u) directly into MeV.
Explain in terms of the Binding Energy per Nucleon curve why both nuclear fusion of hydrogen and nuclear fission of uranium release energy.
Explain the distinct functions of the moderator and the control rods in a thermal nuclear fission reactor.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
An alpha particle (charge +2e, mass 6.64 × 10⁻²⁷ kg) with kinetic energy 5.40 MeV is fired directly head-on at a stationary gold nucleus (¹⁹⁷₇₉Au). Calculate the distance of closest approach of the alpha particle to the centre of the gold nucleus, and calculate the actual nuclear radius of the ¹⁹⁷₇₉Au nucleus given that R₀ = 1.05 × 10⁻¹⁵ m. Explain why the two answers differ. (ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Convert kinetic energy of the alpha particle from MeV to Joules: E_k = 5.40 × 10⁶ eV × 1.60 × 10⁻¹⁹ J eV⁻¹ = 8.64 × 10⁻¹³ J [1 mark]. 2. At the point of closest approach (r_min), the alpha particle comes momentarily to rest and all its kinetic energy is converted into electric potential energy: E_k = (Q_α Q_Au) ÷ (4πε₀ r_min) [1 mark]. 3. Substitute Q_α = 2e and Q_Au = 79e and solve for r_min: r_min = (2 × 79 × (1.60 × 10⁻¹⁹)²) ÷ (4π × 8.85 × 10⁻¹² × 8.64 × 10⁻¹³) = (4.045 × 10⁻³⁶) ÷ (9.609 × 10⁻²³) = 4.21 × 10⁻¹⁴ m (42.1 fm) [1 mark]. 4. Calculate actual nuclear radius of ¹⁹⁷₇₉Au using R = R₀ A^(1/3): R = (1.05 × 10⁻¹⁵ m) × (197)^(1/3) = (1.05 × 10⁻¹⁵) × 5.819 = 6.11 × 10⁻¹⁵ m (6.11 fm) [1 mark]. 5. The alpha particle does not had enough initial kinetic energy (5.40 MeV) to overcome the electrostatic repulsion all the way to the nuclear surface, so it reverses direction at 42.1 fm well outside the 6.11 fm nuclear radius (r_min only gives an upper limit on R) [1 mark].
- Q2
The radioactive isotope Technetium-99m (⁹⁹Tc, molar mass = 0.0990 kg mol⁻¹) is a pure gamma emitter with a half-life of 6.00 hours, widely used as a medical tracer in hospital imaging. A patient is injected with a dose that has an initial activity of 4.80 × 10⁸ Bq. Calculate: (a) the decay constant λ in s⁻¹, (b) the initial mass of Technetium-99m injected into the patient, and (c) the activity remaining in the patient after 24.0 hours (assuming no biological excretion). State two reasons why Technetium-99m is ideal as a medical tracer. (N_A = 6.02 × 10²³ mol⁻¹). (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Convert half-life to seconds: T₁/₂ = 6.00 × 3600 = 21,600 s; calculate decay constant λ = ln(2) ÷ T₁/₂ = 0.6931 ÷ 21,600 = 3.21 × 10⁻⁵ s⁻¹ [1 mark]. 2. (b) Use A₀ = λ N₀ to find the initial number of ⁹⁹Tc nuclei: N₀ = A₀ ÷ λ = (4.80 × 10⁸ Bq) ÷ (3.209 × 10⁻⁵ s⁻¹) = 1.496 × 10¹³ nuclei [1 mark]. 3. Convert N₀ to mass m = (N₀ ÷ N_A) × M_r = (1.496 × 10¹³ ÷ 6.02 × 10²³) × 0.0990 kg mol⁻¹ = 2.46 × 10⁻¹² kg (2.46 ng) [1 mark]. 4. (c) After 24.0 hours (which equals 24.0 ÷ 6.00 = 4 half-lives), Activity A = A₀ × (½)⁴ = (4.80 × 10⁸) ÷ 16 = 3.00 × 10⁷ Bq [1 mark]. 5. Reason 1: It emits only gamma radiation (no strongly ionising alpha or beta particles that would be absorbed inside the body and damage cells), and gamma rays easily pass out of the body to be detected by a gamma camera [1 mark]. 6. Reason 2: Its 6.00-hour half-life is long enough to complete the medical scan with high activity, but short enough that activity decays to safe background levels within a couple of days, minimising the patient's radiation dose (and its daughter product is stable/low-activity) [1 mark].
- Q3
Given the following accurate masses in atomic mass units (u): proton m_p = 1.00728 u, neutron m_n = 1.00867 u, and an Iron-56 (⁵⁶₂₆Fe) nucleus = 55.92067 u (1 u = 931.5 MeV). Calculate the mass defect in u, the total binding energy in MeV, and the binding energy per nucleon in MeV of the ⁵⁶₂₆Fe nucleus. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. An ⁵⁶₂₆Fe nucleus contains Z = 26 protons and (56 - 26) = 30 neutrons [1 mark]. 2. Calculate mass defect Δm = (26 × m_p + 30 × m_n) - m_nucleus = (26 × 1.00728 + 30 × 1.00867) - 55.92067 = (26.18928 + 30.26010) - 55.92067 = 56.44938 - 55.92067 = 0.52871 u [1 mark]. 3. Calculate total binding energy in MeV: BE = 0.52871 u × 931.5 MeV u⁻¹ = 492.5 MeV [1 mark]. 4. Calculate binding energy per nucleon: BE ÷ A = 492.5 MeV ÷ 56 = 8.79 MeV [1 mark].
- Q4
In a nuclear fission reactor, a Uranium-235 nucleus absorbs a thermal neutron and undergoes fission according to the equation: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n. The binding energies per nucleon are: ²³⁵₉₂U = 7.59 MeV, ¹⁴¹₅₆Ba = 8.33 MeV, and ⁹²₃₆Kr = 8.51 MeV. Calculate the energy released in a single fission event in MeV, and calculate the mass of Uranium-235 (molar mass = 0.235 kg mol⁻¹) consumed per day in a nuclear power station that generates a thermal power output of 900 MW. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate total binding energy of reactants (a free neutron has zero binding energy): BE_reactants = 235 × 7.59 MeV = 1783.65 MeV [1 mark]. 2. Calculate total binding energy of products: BE_products = (141 × 8.33) + (92 × 8.51) = 1174.53 + 782.92 = 1957.45 MeV [1 mark]. 3. Energy released per fission event = BE_products - BE_reactants = 1957.45 - 1783.65 = 173.8 MeV (174 MeV) [1 mark]. 4. Convert energy per fission to Joules (173.8 × 1.60 × 10⁻¹³ = 2.781 × 10⁻¹¹ J), and calculate total energy generated in 1 day (86,400 s): E_day = 900 × 10⁶ W × 86,400 s = 7.776 × 10¹³ J [1 mark]. 5. Number of U-235 nuclei fissioned per day N = E_day ÷ E_fission = (7.776 × 10¹³) ÷ (2.781 × 10⁻¹¹) = 2.796 × 10²⁴ nuclei; Mass of U-235 = (N ÷ N_A) × 0.235 = (2.796 × 10²⁴ ÷ 6.02 × 10²³) × 0.235 = 1.09 kg per day [1 mark].
Common Exam Questions (4)
When calculating the mass defect of a nucleus, what do I do if the question gives me the atomic mass of the neutral atom instead of the bare nuclear mass?
If you are given the atomic mass of the neutral atom (which includes the mass of Z orbital electrons), you must either subtract Z × m_e from the atomic mass first to get the bare nuclear mass, OR use the mass of Z neutral Hydrogen-1 atoms (m_H = m_p + m_e) in place of Z protons so the electron masses cancel out automatically.
Why does a single free neutron (¹₀n) have a binding energy of 0 MeV when we use BE values to calculate the energy released in fission, whereas its mass (1.00867 u) must be included when we use atomic masses?
Because a single neutron is just one lone nucleon - it is not bound to any other nucleons, so its binding energy is zero. However, when using actual particle masses (Σm_reactants - Σm_products), a neutron clearly has mass (1.00867 u), so you must count one neutron on the left and 3 neutrons on the right.
How is High-Level, Intermediate-Level, and Low-Level radioactive waste disposed of safely?
1. High-level waste (spent fuel rods containing highly active fission fragments): stored underwater in cooling ponds on-site for several years until decay heat subsides, then vitrified (melted into solid borosilicate glass blocks), sealed in stainless steel canisters, and buried deep underground in geologically stable rock caverns. 2. Intermediate-level waste (reactor casings, fuel cladding): sealed in concrete-filled steel drums. 3. Low-level waste (hospital gloves, syringes): compacted and buried in shallow engineered trenches.
How does carbon-14 dating work, and why can it not be used to date dinosaur bones or rocks?
Cosmic rays convert atmospheric ¹⁴N into radioactive ¹⁴C (half-life = 5730 years), which living plants and animals take in so their ¹⁴C / ¹²C ratio matches the atmosphere. When an organism dies, carbon uptake stops and ¹⁴C decays exponentially (A = A₀ e^(-λt)). Dinosaur fossils (>65 million years old) have undergone >11,000 half-lives so virtually zero ¹⁴C remains (count rate is indistinguishable from background radiation), and rocks were never living organisms.
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