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Radioactivity

CAIE Physics IGCSE: Radioactivity Notes

CAIE Physics IGCSE: Radioactivity Notes

Some isotopes are unstable due to an imbalance of protons and neutrons. They become stable by releasing energy in the form of radiation; this process is called radioactive decay.

5.2 Radioactivity

5.2.1 Detection of Radioactivity

Background radiation is low-level ionising radiation present in the environment from natural and man-made sources:

  • Radon gas from rocks
  • Radiation from food and drink (e.g., radioactive potassium in bananas)
  • Cosmic rays from space
  • Medical and industrial waste

It is detected using a Geiger-Muller (GM) tube connected to a counter.

  • Count rate is the number of decays detected per second (measured in Becquerels, Bq) or per minute.
  • The corrected count rate is found by subtracting the background count rate from the measured count rate:
    Corrected Count Rate = Count Rate with Source - Background Count Rate

5.2.2 The Three Types of Nuclear Emission

Radiation is emitted spontaneously and randomly from the nuclei of unstable isotopes.

Property Alpha (α) Particle Beta (β) Particle Gamma (γ) Ray
Nature Helium nucleus (2p + 2n) High-speed electron Electromagnetic wave
Symbol 42α or 42He 0-1β or 0-1e 00γ
Charge +2 -1 0
Mass 4 1/1840 (negligible) 0
Ionising Power Very High Moderate Very Low
Penetration Low (stopped by paper/skin) Moderate (stopped by few mm aluminium) Very High (stopped by thick lead/concrete)
Deflection by Fields Slightly by electric/magnetic fields Greatly by electric/magnetic fields Not deflected

Rule of Thumb: The more ionising a radiation is, the less penetrating it is.

5.2.3 Radioactive Decay

Radioactive decay is the process by which an unstable nucleus becomes more stable by emitting radiation. It changes the composition of the nucleus.

1. Alpha (α) Decay:

  • The nucleus emits an alpha particle (2 protons + 2 neutrons).
  • Mass number decreases by 4, Atomic number decreases by 2.
  • The atom changes to a different element.
  • Equation:
    AZX → A-4Z-2Y + 42α

2. Beta (β⁻) Decay:

  • A neutron in the nucleus transforms into a proton and emits a beta particle (electron).
  • Mass number stays the same, Atomic number increases by 1.
  • The atom changes to a different element.
  • Equation:
    AZX → AZ+1Y + 0-1β

3. Gamma (γ) Emission:

  • After alpha or beta decay, the nucleus is often left in an excited (high-energy) state. It releases this excess energy as a gamma ray.
  • No change in mass or atomic number. The element remains the same.
  • Equation:
    AZX* → AZX + 00γ
    (The * indicates an excited nucleus)

5.2.4 Nuclear Fission and Fusion

These are processes that involve changes to the nucleus and release vast amounts of energy, much of it in the form of radiation.

Nuclear Fission:

  • The splitting of a large, unstable nucleus into two smaller, more stable nuclei.
  • Induced when a free neutron is absorbed by the nucleus (e.g., Uranium-235 or Plutonium-239).
  • Releases 2 or 3 more neutrons and a large amount of energy (as gamma radiation and kinetic energy).
  • These new neutrons can go on to cause more fission events, leading to a chain reaction.
  • Used in nuclear power plants and atomic weapons.
  • Fission Equation Example:
    23592U + 10n → 9236Kr + 14156Ba + 310n + energy

Nuclear Fusion:

  • The joining of two light nuclei to form a heavier nucleus.
  • Requires extremely high temperatures and pressures to overcome the electrostatic repulsion between positively charged nuclei.
  • The mass of the product nucleus is less than the mass of the original nuclei; the lost mass is converted into a huge amount of energy (primarily as gamma radiation).
  • This is the process that powers stars, including the Sun.
  • Fusion Equation Example (The Sun):
    21H + 31H → 42He + 10n + energy

Key Rule for All Nuclear Equations:

  • The sum of the mass numbers (A) must be equal on both sides of the equation.
  • The sum of the atomic numbers (Z) must be equal on both sides of the equation.

5.2.5 Half-Life

The half-life of a radioactive isotope is the time taken for half of the radioactive nuclei in a sample to decay, or for its activity (count rate) to fall by half.

  • It is constant for a given isotope.
  • To find it from a graph, find the time for the count rate to drop from any value to half of that value.
  • Always subtract the background count before performing half-life calculations.

Applications of Radioisotopes:

  • Smoke Alarms: Use a long-half-life α-emitter (e.g., Americium-241). Smoke absorbs alpha particles, reducing the current and triggering the alarm.
  • Medical Tracers & Therapy: Use short-half-life γ-emitters (e.g., Technetium-99m) for imaging. Use intense γ-emitters (e.g., Cobalt-60) to kill cancer cells.
  • Sterilisation & Food Irradiation: Use γ-rays to kill bacteria on medical equipment or in food without damaging the packaging.
  • Thickness Control: Use β-emitters to monitor the thickness of sheets of material (e.g., paper, metal) during production.

5.2.6 Safety Precautions

Ionising radiation can damage or kill cells and cause mutations (cancer) by breaking DNA molecules.

Safety Protocols:

  • Minimize Time: Spend as little time as possible near a source.
  • Maximize Distance: Use tongs; keep the source at arm's length. Intensity follows the inverse square law.
  • Use Shielding:
    • α-particles: Paper, gloves.
    • β-particles: Aluminium.
    • γ-rays: Thick lead or concrete.
  • Safe Storage: Keep sources in a lead-lined box when not in use.
  • Disposal: Use isotopes with a short half-life where possible so they quickly become safe.

Abel Masitsa.

Why alpha particles are more ionising but less penetrating

A clear, exam-friendly note for IGCSE (includes the mark-scheme explanation and the deeper physics).

Key facts

  • Alpha particle = helium nucleus (2 protons + 2 neutrons). Charge: +2. Mass: large.
  • Beta particle = electron. Charge: −1 (or +1 for positron). Mass: very small.
  • Gamma ray = electromagnetic radiation (no charge, no rest mass).

Why alpha particles are more ionising

Alpha particles cause more ionisation because they are heavy and carry a +2 charge. These properties make them interact strongly with atoms, knocking out many electrons per unit distance travelled. In many radioactive decays alphas also have larger kinetic energy than betas, which further increases energy transfer per collision.

Why alpha particles are less penetrating

Because alphas lose energy rapidly through intense ionisation, they cannot travel far. A sheet of paper or the outer dead layer of skin will stop them. Betas are lighter and interact less strongly, so they penetrate further than alphas but less than gammas.

Quick kinetic energy check (short)

Kinetic energy: E_k = ½ m v². Alphas have a very large mass; even if their speed is ≈ 1/20 of a beta, the product m v² can be larger for alphas. Thus alphas often have more KE than betas in real decays — this supports Cambridge’s simplified phrasing that alphas are more ionising because they have more kinetic energy.

Exam tip (IGCSE)

If the mark scheme gives the statement linking ionising power to kinetic energy, choose the option that matches the mark scheme. In your own answer you can add that mass and charge are the fundamental reasons for high ionisation by alpha particles.

Summary table

Radiation Mass / charge Ionising power Penetration
Alpha Large mass, +2 High Low (stopped by paper/skin)
Beta Very small mass, −1/+1 Lower than alpha Medium (stopped by thin metal)
Gamma No mass, no charge Lowest Highest (requires thick lead/concrete)

Short summary: Alpha → more ionising, less penetrating; Beta → less ionising, more penetrating; Gamma → least ionising, most penetrating.

How Smoke/Fire alarms work

In ionization smoke alarms, alpha particles from a small amount of americium-241 are used.

Here’s how it works:

  • The americium-241 gives off alpha radiation, which ionizes the air molecules in a small chamber inside the alarm.

  • This creates a flow of ions (positive and negative charges), allowing a small electric current to pass between two electrodes.

  • When smoke enters the chamber, it attaches to the ions and reduces their mobility, which disrupts the current.

  • The alarm senses this drop in current and triggers the warning sound.

In short:

👉 In smoke alarms, alpha particles are used to ionize the air, allowing current to flow in the circuit.

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Half-life formulas — quick guide

Half-life formulas — quick reference

1. Exponential decay (use when you know the half-life)

\(N = N_0 \left(\tfrac{1}{2}\right)^{\tfrac{t}{T_{1/2}}}\)
Meaning: Remaining amount N after time t, starting from N0. T1/2 is the half-life.
Example (quick)
Start: 80 g, half-life 5 days, t = 12 days. Number of half-lives n = 12/5 = 2.4. Remaining: 80*(1/2)^{2.4} \approx 80*0.189 = 15.12 g.

2. Fraction remaining

\(\text{Fraction remaining} = \left(\tfrac{1}{2}\right)^{\tfrac{t}{T_{1/2}}}\)
Use this when you only need a fraction or percentage of the original sample after time t.

3. Number of half-lives

\(n = \dfrac{t}{T_{1/2}}\)
If n is an integer, the sample halves exactly that many times.

4. Solve for half-life (when you know N, N0, and t)

\(T_{1/2} = \dfrac{t\,\ln 2}{\ln\left(\dfrac{N_0}{N}\right)}\)
Derived by taking natural logs of the exponential decay law.
Worked example
Suppose N0=200 counts, N=50 counts after t = 9 hours. \(\dfrac{N_0}{N}=4\) so \(\ln(4)=1.386...\). Then \(T_{1/2}=\dfrac{9\times0.6931}{1.386}=\dfrac{6.238}{1.386}=4.5\) hours.

5. Decay constant and exponential form

\(\lambda = \dfrac{\ln 2}{T_{1/2}},\quad N = N_0 e^{-\lambda t}\)
Use this when the question uses the exponential constant \(\lambda\) instead of half-life.

Quick step-by-step pattern for solving problems

  1. Write down what you know: N, N0, t, or T1/2.
  2. Choose the formula above that contains the unknown.
  3. If needed, take natural logs to solve for the unknown; show units.
  4. Check special cases: if t = T1/2, remaining fraction = 1/2; if t = 0, remaining = N0.

Worked example (full)

Problem: A 160 g sample decays to 10 g in 30 days. Find the half-life.

Use formula: T_{1/2} = t*ln2 / ln(N0/N)
Substitute: t = 30, N0 = 160, N = 10
N0/N = 16, ln(16) = 2.7726
T_{1/2} = 30 * 0.6931 / 2.7726 = 20.79 / 2.7726 = 7.5 days

Answer: T_{1/2} = 7.5 days
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