43  Properties of the Nucleus

43.1 Syllabus inquiry question

  • How can the nucleus be transformed?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 4:

The nucleus is held together by a force much stronger than electricity—the nuclear force. It acts only over very short distances, but within that range it dominates. This is why so much energy is released when nuclei change.

43.2 Learning Objectives

  • Analyse radioactive decay: alpha, beta, and gamma
  • Apply the exponential decay equation: \(N_t = N_0 e^{-\lambda t}\)
  • Calculate half-life and decay constant: \(\lambda = \ln 2/t_{1/2}\)
  • Analyse mass defect and binding energy: \(E = mc^2\)
  • Compare energy release in fission and fusion

43.3 Content

43.3.1 Radioactive Decay

Radioactive decay is the spontaneous emission of radiation from unstable nuclei.

Type Particle Symbol Charge Mass Penetration
Alpha Helium nucleus \(^4_2\alpha\) +2 4 u Paper stops it
Beta⁻ Electron \(^0_{-1}\beta\) -1 ~0 Few mm aluminium
Beta⁺ Positron \(^0_{+1}\beta\) +1 ~0 Few mm aluminium
Gamma Photon γ 0 0 Lead/concrete

43.3.2 Interactive: Radioactive Decay Types

43.3.3 Decay Equations

Alpha decay: Nucleus loses 2 protons and 2 neutrons \[^A_Z X \rightarrow ^{A-4}_{Z-2} Y + ^4_2\alpha\]

Beta⁻ decay: Neutron → proton + electron + antineutrino \[^A_Z X \rightarrow ^{A}_{Z+1} Y + ^0_{-1}\beta + \bar{\nu}_e\]

Beta⁺ decay: Proton → neutron + positron + neutrino \[^A_Z X \rightarrow ^{A}_{Z-1} Y + ^0_{+1}\beta + \nu_e\]

Gamma emission: Nucleus loses energy without changing composition \[^A_Z X^* \rightarrow ^{A}_{Z} X + \gamma\]

Conservation Laws

In all nuclear reactions: - Mass number (A) is conserved - Atomic number (Z) is conserved - Charge is conserved - Energy is conserved (including mass-energy)

43.3.4 Exponential Decay

The number of undecayed nuclei decreases exponentially:

\[N_t = N_0 e^{-\lambda t}\]

where: - \(N_t\) = number of nuclei remaining at time t - \(N_0\) = initial number of nuclei - \(\lambda\) = decay constant (s⁻¹) - \(t\) = time (s)

43.3.5 Interactive: Exponential Decay

43.3.6 Half-Life

Half-life (\(t_{1/2}\)) is the time for half the nuclei to decay:

\[\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}\]

After Fraction Remaining
0 half-lives 1 (100%)
1 half-life 1/2 (50%)
2 half-lives 1/4 (25%)
3 half-lives 1/8 (12.5%)
n half-lives 1/2ⁿ

43.3.7 Activity

Activity (A) is the number of decays per second:

\[A = \lambda N = A_0 e^{-\lambda t}\]

Unit: Becquerel (Bq) = 1 decay per second

43.3.8 Interactive: Activity vs Time

43.3.9 Mass Defect

The mass defect is the difference between: - Mass of separated nucleons (protons + neutrons) - Actual mass of the nucleus

\[\Delta m = Zm_p + (A-Z)m_n - m_{nucleus}\]

This “missing mass” has been converted to binding energy.

43.3.10 Interactive: Mass Defect

43.3.11 Binding Energy

Binding energy is the energy required to separate all nucleons:

\[E_b = \Delta m \cdot c^2\]

Binding energy per nucleon indicates nuclear stability:

\[\frac{E_b}{A} = \frac{\Delta m \cdot c^2}{A}\]

43.3.12 Interactive: Binding Energy Curve

43.3.13 Key Features of the Binding Energy Curve

Feature Significance
Peak at Fe-56 Most stable nucleus (~8.8 MeV/nucleon)
Light nuclei (left) Fusion releases energy (moving toward peak)
Heavy nuclei (right) Fission releases energy (moving toward peak)
H-2, H-3, He-3 Low BE/A; unstable or fusion fuel
Energy Release

Both fusion (of light nuclei) and fission (of heavy nuclei) release energy because the products are more tightly bound (higher BE/A) than the reactants.

43.3.14 Nuclear Fission

Fission is the splitting of a heavy nucleus into lighter nuclei:

\[^{235}_{92}\text{U} + ^1_0\text{n} \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3^1_0\text{n} + \text{energy}\]

Key features: - Triggered by neutron capture - Releases ~200 MeV per fission - Produces 2-3 neutrons → chain reaction possible - Mass defect → energy release

43.3.15 Nuclear Fusion

Fusion is the combining of light nuclei into heavier nuclei:

\[^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n} + 17.6\ \text{MeV}\]

Key features: - Requires extreme temperature (~10⁷ K) to overcome Coulomb repulsion - Powers the Sun and stars - Produces no long-lived radioactive waste - Mass defect → energy release

43.3.16 Interactive: Fission vs Fusion

43.3.17 Applications of Nuclear Physics

Application Type Use
Nuclear power Fission Electricity generation
Nuclear weapons Fission/Fusion Military
Medical isotopes Various PET scans, cancer treatment
Carbon dating β decay Archaeology
Smoke detectors α decay Fire safety

43.4 Worked Examples

43.4.1 Example 1: Half-life decay

A sample contains 8.0 × 10⁶ radioactive nuclei. The half-life is 2.0 hours. How many remain after 6.0 hours?

Solution:

  1. Number of half-lives: \(n = 6.0/2.0 = 3\)

  2. Fraction remaining: \(1/2^3 = 1/8\)

  3. \(N_t = 8.0 \times 10^6 / 8 = 1.0 \times 10^6\) nuclei

43.4.2 Example 2: Decay constant

A radioactive isotope has half-life 5.0 hours. Calculate the decay constant.

Solution:

  1. Use \(\lambda = \ln 2/t_{1/2}\)

  2. \(\lambda = 0.693/(5.0 \times 3600)\)

  3. \(\lambda = 3.85 \times 10^{-5}\ \text{s}^{-1}\)

43.4.3 Example 3: Exponential decay equation

Using the decay constant from Example 2, calculate the fraction remaining after 10 hours.

Solution:

  1. Use \(N_t/N_0 = e^{-\lambda t}\)

  2. \(t = 10 \times 3600 = 36000\) s

  3. \(N_t/N_0 = e^{-3.85 \times 10^{-5} \times 36000} = e^{-1.39} = 0.25\)

(This is 2 half-lives, so 1/4 remaining—confirms our answer)

43.4.4 Example 4: Mass defect and binding energy

Calculate the binding energy of He-4 given: mass of He-4 = 4.0026 u, mp = 1.0073 u, mn = 1.0087 u.

Solution:

  1. Mass of separated nucleons: \(2(1.0073) + 2(1.0087) = 4.0320\) u

  2. Mass defect: \(\Delta m = 4.0320 - 4.0026 = 0.0294\) u

  3. Convert to kg: \(\Delta m = 0.0294 \times 1.66 \times 10^{-27} = 4.88 \times 10^{-29}\) kg

  4. \(E = \Delta m c^2 = 4.88 \times 10^{-29} \times (3.0 \times 10^8)^2 = 4.4 \times 10^{-12}\) J = 27.5 MeV

43.4.5 Example 5: Energy from fission

If 1.0 kg of U-235 undergoes fission, releasing 200 MeV per nucleus, calculate the total energy released.

Solution:

  1. Number of nuclei: \(N = (1.0 \times 10^3)/(235 \times 1.66 \times 10^{-27}) = 2.56 \times 10^{24}\)

  2. Energy per nucleus: \(200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\) J

  3. Total energy: \(E = 2.56 \times 10^{24} \times 3.2 \times 10^{-11} = 8.2 \times 10^{13}\) J

(This is equivalent to ~20 kilotons of TNT!)

43.5 Common Misconceptions

Common Misconceptions
  • Misconception: Half-life means half the sample has physically disappeared. Correction: Half the radioactive nuclei have transformed into different nuclei. The atoms are still there, just changed.

  • Misconception: After 2 half-lives, all nuclei have decayed. Correction: After 2 half-lives, 1/4 remain. After many half-lives, some nuclei still remain (exponential decay never reaches zero).

  • Misconception: Fission releases more energy per nucleon than fusion. Correction: Fusion releases ~4× more energy per nucleon. Fission releases more total energy per reaction because uranium nuclei are so much heavier.

  • Misconception: Binding energy is the energy stored in the nucleus. Correction: Binding energy is the energy required to separate the nucleus. A higher binding energy means a more stable, lower-energy nucleus.

  • Misconception: Nuclear reactions convert mass entirely into energy. Correction: Only a small fraction of mass (the mass defect) is converted. Most mass remains as matter.

43.6 Practice Questions

43.6.1 Easy (2 marks)

A radioactive sample has a half-life of 4.0 days. Calculate the decay constant in s⁻¹.

  • Use \(\lambda = \ln 2/t_{1/2}\) (1)
  • \(\lambda = 0.693/(4.0 \times 24 \times 3600) = 2.0 \times 10^{-6}\) s⁻¹ (1)

Answer: λ = 2.0 × 10⁻⁶ s⁻¹

43.6.2 Medium (4 marks)

A sample initially contains 6.0 × 10⁸ nuclei with decay constant 0.25 h⁻¹. Calculate the number remaining after 8.0 hours and the half-life.

  • \(N_t = N_0 e^{-\lambda t} = 6.0 \times 10^8 \times e^{-0.25 \times 8}\) (1)
  • \(N_t = 6.0 \times 10^8 \times e^{-2} = 6.0 \times 10^8 \times 0.135\) (1)
  • \(N_t = 8.1 \times 10^7\) nuclei (1)
  • \(t_{1/2} = \ln 2/\lambda = 0.693/0.25 = 2.77\) hours (1)

Answer: 8.1 × 10⁷ nuclei; t₁/₂ = 2.8 hours

43.6.3 Hard (5 marks)

Calculate the energy released when 1 mole of deuterium-tritium fusion occurs: D + T → He-4 + n. Masses: D = 2.0141 u, T = 3.0160 u, He = 4.0026 u, n = 1.0087 u.

  • Total reactant mass: 2.0141 + 3.0160 = 5.0301 u (1)
  • Total product mass: 4.0026 + 1.0087 = 5.0113 u (1)
  • Mass defect: Δm = 5.0301 - 5.0113 = 0.0188 u per reaction (1)
  • Energy per reaction: E = 0.0188 × 931.5 = 17.5 MeV (1)
  • Energy per mole: 17.5 × 6.02 × 10²³ = 1.05 × 10²⁵ MeV = 1.7 × 10¹² J (1)

Answer: 17.5 MeV per reaction; 1.7 × 10¹² J per mole

43.7 Multiple Choice Questions

Test your understanding with these interactive questions:

43.8 Summary

Key Takeaways
  • Alpha (α): \(^4_2\)He nucleus; beta (β): electron/positron; gamma (γ): photon
  • Exponential decay: \(N_t = N_0 e^{-\lambda t}\)
  • Half-life: \(t_{1/2} = \ln 2/\lambda\)
  • Mass defect: \(\Delta m = \sum m_{nucleons} - m_{nucleus}\)
  • Binding energy: \(E_b = \Delta m c^2\)
  • Fission: heavy → lighter nuclei + energy
  • Fusion: light → heavier nuclei + energy
  • Both release energy by moving toward Fe-56 on binding energy curve

43.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: