understand that isotopes are forms of the same element with different numbers of neutrons in their nuclei

Topic: Atoms, Nuclei and Radiation (Cambridge IGCSE/A‑Level 9702)

Learning Objective

Explain that isotopes are different forms of the same element that have the same number of protons (Z) but different numbers of neutrons (N), giving different mass numbers (A). Extend this understanding to nuclear stability, radioactive decay, conservation laws, mass‑defect, binding energy and the fundamental particles that make up nuclei.

1. Basic Nuclear Notation & Simple Nuclear Model

• Atomic number (Z): number of protons; defines the chemical element.
• Neutron number (N): N = A – Z.
• Mass number (A): total number of nucleons (protons + neutrons).
• Isotope notation: \$_{Z}^{A}\text{X}\$, where X is the element symbol.

Simple nuclear model (Rutherford/Bohr)

The nucleus is a tiny, massive core containing all the protons and neutrons. Electrons occupy the surrounding space and are treated as point‑like particles orbiting a point‑like nucleus. The nucleus occupies <≈10⁻⁵ % of the atom’s volume.

Rutherford α‑particle scattering experiment

Rutherford directed a beam of α‑particles at a thin gold foil. Most particles passed straight through, but a few were deflected at large angles. This could only be explained if the positive charge (and most of the mass) were concentrated in a very small, dense nucleus, confirming the nuclear model.

2. Isotopes – Definition, Notation, Stability and Applications

Physical consequences of different neutron numbers

1. Mass differences change the element’s average atomic mass.
2. Unstable isotopes undergo radioactive decay to reach a more stable configuration.
3. Isotopic composition has only a marginal effect on chemical behaviour, but is crucial for dating, medical imaging and nuclear reactors.

3. Radioactive Decay – Types, Equations and Conservation Laws

3.1 Decay modes

3.2 Conservation in nuclear reactions

• Charge (Z) is conserved. The sum of atomic numbers before and after a reaction is identical.
• Mass number (A) is conserved. The total number of nucleons (including emitted particles) remains the same.
• Energy and momentum are also conserved; the Q‑value represents the net energy released.

3.3 Worked example – β⁻ decay of ⁶⁰Co

Write the complete nuclear equation, showing the change in both Z and A:

\[

{27}^{60}\text{Co} \;\rightarrow\; {28}^{60}\text{Ni} \;+\; e^- \;+\; \bar{\nu}_e \quad (Q \approx 2.82\;\text{MeV})

\]

Explanation:

• Neutron in the cobalt nucleus converts to a proton, increasing Z from 27 to 28 while A stays 60.
• An electron (β⁻ particle) and an antineutrino are emitted to conserve charge and lepton number.

3.4 Decay series

Heavy nuclei often decay through a chain of α and β steps until a stable nuclide is reached. Example: the ²³⁸U series ends in ²⁰⁶Pb.

4. Quantitative Aspects of Radioactivity

4.1 Half‑life and decay constant

\[

t_{1/2} = \frac{0.693}{\lambda}, \qquad

N(t)=N_0 e^{-\lambda t}, \qquad

A = \lambda N

\]

• λ = decay constant (s⁻¹)
• t½ = half‑life
• N = number of undecayed nuclei
• A = activity (Bq = decays s⁻¹)

Sample calculation 1 – Activity of 1 g ¹⁴C

1. Atomic mass ≈ 14.003 u → 1 mol = 14.003 g.
2. Number of atoms:

   \[

   N = \frac{1\;\text{g}}{14.003\;\text{g mol}^{-1}} \times N_A

   = 4.29\times10^{22}\;\text{atoms}

   \]

3. Half‑life \$t{1/2}=5730\;\$yr \$=1.81\times10^{11}\;\$s → \$\lambda = 0.693/t{1/2}=3.83\times10^{-12}\;\$s⁻¹.
4. Activity:

   \[

   A = \lambda N = 3.83\times10^{-12}\times4.29\times10^{22}

   \approx 1.64\times10^{11}\;\text{Bq}

   \]

   ≈ 164 GBq.

Sample calculation 2 – Binding energy of ⁴He

• Mass of ⁴He nucleus = 4.002603 u.
• Mass of constituent nucleons:

  \[

  2p + 2n = 2(1.007276\;\text{u}) + 2(1.008665\;\text{u}) = 4.031882\;\text{u}

  \]

• Mass defect:

  \[

  \Delta m = 4.031882 - 4.002603 = 0.029279\;\text{u}

  \]

• Binding energy:

  \[

  E = \Delta m \, c^{2} = 0.029279\;\text{u}\times931.5\;\frac{\text{MeV}}{\text{u}}

  \approx 27.2\;\text{MeV}

  \]

• Binding energy per nucleon = 27.2 MeV / 4 ≈ 6.8 MeV.

4.3 Mass‑defect and binding‑energy curve

The curve of binding energy per nucleon versus mass number peaks near ⁵⁶Fe (≈ 8.8 MeV per nucleon). Nuclei lighter than Fe release energy by fusion; heavier nuclei release energy by fission.

5. Fundamental Particles Relevant to Nuclear Physics

5.1 Quarks and hadrons

• Quarks (six flavours): up (u), down (d), strange (s), charm (c), bottom (b), top (t).
  

  Charge: u = +2/3 e, d = ‑1/3 e (others analogous).

• Baryons = three‑quark states (e.g., proton = uud, neutron = udd).
  

  Baryon number \$B = 1\$.

• Mesons = quark–antiquark pairs (e.g., \$\pi^{+}=u\bar d\$).
  

  Baryon number \$B = 0\$.

5.2 Leptons

5.3 Antiparticles

Every particle has a corresponding antiparticle with opposite charge (e.g., proton ↔ antiproton, electron ↔ positron). In β⁺‑decay a positron (e⁺) is emitted and a neutrino (νₑ) is produced; in β⁻‑decay an electron (e⁻) and an antineutrino (\$\barν_e\$) are emitted.

5.4 Role of neutrinos

• Carry away the “missing” energy and momentum in β‑decay, ensuring conservation of energy, momentum and lepton number.
• Interact only via the weak force, so they escape most detectors unnoticed.

6. Practical Skills (AO3)

6.1 Designing a simple decay experiment

1. Choose a radioactive source with a known half‑life (e.g., ⁶⁰Co).
2. Set up a Geiger‑Müller tube connected to a data logger.
3. Record counts per minute (cpm) at regular intervals (e.g., every 5 min) for several half‑lives.
4. Convert cpm to activity (Bq) using the detector efficiency.
5. Plot \$\ln\$(activity) versus time; the slope equals \$-λ\$.
6. Calculate \$t_{1/2}=0.693/λ\$ and compare with the accepted value.

6.2 Sample calculation – Determining activity from half‑life

Given \$t_{1/2}=5.27\,\$yr for ⁶⁰Co, find the activity of a 2 g sample.

1. Atomic mass ≈ 60 u → 1 mol = 60 g.
2. Number of atoms in 2 g:

   \$N = \dfrac{2}{60}\,N_A = 2.01\times10^{22}\$.

3. λ = \$0.693 / (5.27\;\text{yr}\times3.16\times10^{7}\;\text{s yr}^{-1}) = 4.16\times10^{-9}\;\$s⁻¹.
4. Activity \$A = λN = 8.36\times10^{13}\;\$Bq ≈ 83 TBq.

6.3 Calculating average atomic mass

For chlorine, natural abundances are 75.78 % ³⁵Cl and 24.22 % ³⁷Cl.

\[

\bar{M} = (0.7578)(34.969) + (0.2422)(36.966) = 35.45\;\text{u}

\]

7. Common Misconceptions & Corrections

• Misconception: Isotopes are different elements.
  

  Correction: They have the same \$Z\$ (same element) but different \$N\$ and therefore different \$A\$.

• Misconception: The nucleus is the same size as the atom.
  

  Correction: The nucleus occupies only about 10⁻⁵ % of the atomic volume; the electron cloud defines the atom’s size.

• Misconception: Mass number is conserved but charge is not in decay.
  

  Correction: Both \$A\$ and \$Z\$ are conserved in any nuclear reaction; emitted particles carry the necessary changes.

• Misconception: β‑decay violates conservation of energy.
  

  Correction: The antineutrino (or neutrino) carries away the missing energy, preserving energy and momentum.