Ionizing radiation and risk

 Module 6: Field and particle physics

6.2 Fundamental particle

6.2.2 Ionizing radiation and risk

a)   Describe and explain:

I)                   The nature and effects of ionizing radiations: differences in ionizing and penetrating power, effects on living tissue

II)                The stability and decay of nuclei in terms of binding energy; transformation of nucleus on emission of radiation; qualitative variation of binding energy with proton and neutron number (“Nuclear Valley”)

III)             Nuclear fission; chain reaction; nuclear fusion; nuclear power generation.

b)   Make appropriate use of:

I)                   The terms: nucleon number, proton number, isotope, binding energy, atomic mass unit, absorbed and effective dose, risk

by sketching and interpreting:

II)                Plots of binding energy per nucleon of nuclei against nucleon number.

c)    Make calculations and estimates involving:

I)                   Activity of a sample of radioactive material (related to half-life or decay constant)

II)                Absorbed dose in gray = energy deposited per unit mass

III)             Effective dose in sievert = absorbed dose in gray × quality factor

IV)             Energy changes from nuclear transformations: [math]E_{rest} = mc^2[/math]

d)   Demonstrate and apply knowledge and understanding of the following practical activities (HSW4):

I)                   Studying the absorption of α-particles, β-particles and γ-rays by appropriate materials

II)                Determining the half-life of an isotope such as protactinium.

  • a) Describe and explain:

  • I) The Nature and Effects of Ionizing Radiations

  • ⇒ Nature of Ionizing Radiation
  • Definition: Ionizing radiation refers to energy in the form of particles or electromagnetic waves that has enough energy to ionize atoms by knocking electrons off them.
  • ⇒ Types of Ionizing Radiation:
  • Alpha (α) particles:
  • – Helium nuclei ([math]{}_2^4\text{He}[/math]) emitted from unstable nuclei. Heavy and positively charged.
  • Beta (β) particles:
  • – High-energy electrons ([math]β^-[/math]) or positrons ([math]β^+[/math]) emitted during nuclear decay.
  • Gamma (γ) rays: High-energy electromagnetic radiation emitted from a nucleus after radioactive decay
  • X-rays: High-energy electromagnetic radiation generated artificially or by interactions in an atom.
  • Neutrons: Neutral particles emitted during some nuclear reactions, such as fission.
  • Figure 1 Ionizing Radiation
  • ⇒  Differences in Ionizing and Penetrating Power
  • 1. Alpha particles:
  • – Ionizing Power: Very high due to large mass and charge.
  • – Penetrating Power: Very low; can be stopped by a sheet of paper or skin.
  • Figure 2 Alpha Particle
  • 2. Beta particles:
  • – Ionizing Power: Moderate; lighter and less charged than alpha particles.
  • – Penetrating Power: Moderate; can penetrate paper but stopped by thin metal (e.g., aluminum).
  • Figure 3 Beta particles
  • 3. Gamma rays:
  • – Ionizing Power: Low compared to particles.
  • – Penetrating Power: Very high; requires thick lead or concrete for shielding.
  • Figure 4 Gamma rays
  • 4. Neutrons:
  • – Ionizing Power: Indirect (causes secondary ionization through collisions).
  • – Penetrating Power: High; requires hydrogen-rich materials (e.g., water or polyethylene) for absorption.
  • ⇒ Effects on Living Tissue
  • 1. DNA Damage: Radiation can break DNA strands or cause mutations, leading to cancer or cell death.
  • 2. Acute Effects:
  • – Low doses: Increased risk of cancer.
  • – High doses: Radiation burns, sickness (nausea, fatigue), and organ failure.
  • 3. Chronic Effects: Long-term exposure can lead to genetic mutations, sterility, or cataracts.
  • 4. Deterministic vs Stochastic Effects:
  • – Deterministic: Severity increases with dose (e.g., radiation burns).
  • – Stochastic: Probability increases with dose (e.g., cancer).

  • II) Stability and Decay of Nuclei

  • ⇒  Binding Energy and Stability
  • 1. Binding Energy:
  • – The energy required to separate a nucleus into its protons and neutrons.
  • – Greater binding energy per nucleon indicates a more stable nucleus.
  • 2. Nuclear Stability:
  • – Stable nuclei have an optimal ratio of protons to neutrons.
  • – Too many or too few neutrons cause instability, leading to radioactive decay.
  • ⇒  Nuclear Decay and Transformation
  • 1. Alpha Decay:
  • – Emission of an [math]{}_2^4\text{He}[/math] nucleus
  • – Decreases atomic number by 2 and mass number by 4.
  • – Example:
  • [math]{}_{92}^{238}\text{U} \rightarrow {}_{90}^{234}\text{Th} + \alpha[/math]
  • 2. Beta Decay:
  • – Beta-minus ([math]β^-[/math]): Neutron converts to a proton, emitting an electron and antineutrino.
  • – Example:
  • [math]{}_{6}^{14}\text{C} \rightarrow {}_{7}^{14}\text{N} + \beta^- + \bar{\nu}_e[/math]
  • – Beta-plus ([math]β^+[/math]):
  • – Proton converts to a neutron, emitting a positron and neutrino.
  • 3. Gamma Decay:
  • – Nucleus emits excess energy as gamma rays without changing its composition.
  • ⇒  Nuclear Valley (Binding Energy Curve)
  • 1. Qualitative Variation:
  • – Binding energy per nucleon peaks at iron ([math]Fe^{56}[/math]), indicating maximum stability.
  • – Lighter nuclei (e.g., hydrogen) can release energy through fusion.
  • – Heavier nuclei (e.g., uranium) release energy through fission.
  • 2. Nuclear Valley:
  • – Visual representation of nuclear stability shows a “valley” with iron at the bottom, representing the most stable nuclei.

  • III) Nuclear Fission, Fusion, and Power Generation

  • ⇒ Nuclear Fission
  • 1. Definition: Splitting of a heavy nucleus into two smaller nuclei, releasing energy.
  • 2. Process:
  • – Initiated by neutron absorption
  • – Example:
  • [math]{}_{92}^{235}\text{U} + n \rightarrow {}_{92}^{236}\text{U} \rightarrow {}_{36}^{92}\text{Kr} + {}_{56}^{141}\text{Ba} + 3n + \text{Energy}[/math]
  • Figure 5 Nuclear Fission
  • 3. Chain Reaction:
  • – Released neutrons induce further fission events.
  • – Controlled in nuclear reactors; uncontrolled in nuclear weapons.
  • 4. Energy Source:
  • – Fission releases a significant amount of energy (from binding energy differences).
  • ⇒ Nuclear Fusion
  • 1. Definition: Combining two light nuclei to form a heavier nucleus, releasing energy.
  • 2. Process:
  • – Requires extremely high temperatures and pressures to overcome electrostatic repulsion.
  • – Example:
  • [math]{}_{1}^{2}\text{H} + {}_{1}^{3}\text{H} \rightarrow {}_{2}^{4}\text{He} + n + \text{Energy}[/math]
  • Figure 6 Nuclear Fusion
  • 3. Energy Potential:
  • Fusion releases more energy per reaction than fission.
  • Powers stars, including the Sun.
  • ⇒  Nuclear Power Generation
  • 1. Fission Reactors:
  • – Use controlled chain reactions to produce heat.
  • – Heat is used to generate steam, driving turbines to produce electricity.
  • 2. Fusion Reactors (Experimental):
  • – Aim to replicate star-like conditions on Earth.
  • – Challenges include sustaining high temperatures and pressures and achieving net positive energy.

  • b) Make appropriate use of:

  • I) Terms and Definitions

  • 1. Nucleon Number (Mass Number):
  • – Total number of protons and neutrons in a nucleus.
  • – Denoted as A: A = Z + N, where Z is the proton number, and N is the neutron number.
  • – Example:
  • [math]\text{Carbon-12 } ({}_{6}^{12}\text{C})[/math] A = 12.
  • 2. Proton Number (Atomic Number):
  • – Number of protons in the nucleus of an atom.
  • – Determines the chemical element.
  • – Denoted as Z.
  • – Example: Carbon has Z = 6, meaning it has 6 protons.
  • 3. Isotope:
  • – Atoms of the same element with the same number of protons (Z) but different numbers of neutrons (N)
  • – Example: Hydrogen isotopes include:
  • – Protium ([math]{}_{1}^{1}\text{H}[/math]): 1 proton, 0 neutrons.
  • – Deuterium ([math]{}_{1}^{2}\text{H}[/math]): 1 proton, 1 neutron.
  • – Tritium ([math]{}_{1}^{3}\text{H}[/math]): 1 proton, 2 neutrons.
  • 4. Binding Energy:
  • – The energy required to disassemble a nucleus into its individual protons and neutrons.
  • – It represents the “glue” holding the nucleus together.
  • – Measured in mega-electron volts (MeV).
  • 5. Atomic Mass Unit (u):
  • – Unit of mass used to express atomic and nuclear masses.
  • – Defined as [math]1u = \frac{1}{12} [/math] the mass of a carbon-12 atom
  • – [math]1u ≈ 1.660 × 10^{-27} kg[/math]
  • 6. Absorbed Dose:
  • – The amount of radiation energy absorbed per unit mass of tissue.
  • – Measured in grays (Gy), where 1Gy=1J/kg.
  • 7. Effective Dose:
  • – Accounts for the type of radiation and the sensitivity of tissues to radiation.
  • – Measured in sieverts (Sv).
  • – Effective Dose=Absorbed Dose × Radiation Weighting Factor
  • 8. Risk:
  • – The probability of harm (e.g., cancer, genetic mutations) caused by exposure to radiation.
  • – Quantified based on dose, duration, and type of radiation.

  • II) Plot of Binding Energy per Nucleon vs. Nucleon Number

  • ⇒  Sketch:
  • The graph of binding energy per nucleon ([math]\frac{E_b}{A}[/math]) nucleon number (A) has the following characteristics
  • Horizontal Axis (A): Nucleon number (mass number), ranging from light nuclei (A≈1) to heavy nuclei (A>200).
  • Vertical Axis ([math]\frac{E_b}{A}[/math]): Binding energy per nucleon, measured in MeV.
  • ⇒  Features of the Graph:
  • 1. Light Nuclei (A<20):
  • – Binding energy per nucleon increases rapidly as nucleon number increases.
  • – Example: Hydrogen ([math]{}_{1}^{1}\text{H}[/math]) has [math]\frac{E_b}{A}[/math]; Helium-4 ([math]{}_{2}^{4}\text{H}[/math]) has [math]\frac{E_b}{A} ≈ 7MeV[/math].
  • 2. Intermediate Nuclei (A≈20 to 60):
  • – Binding energy per nucleon peaks around A=56 (Iron-56), with [math]\frac{E_b}{A} ≈ 8.8MeV[/math], the most stable nucleus.
  • – Elements in this range are highly stable due to strong nuclear forces.
  • 3. Heavy Nuclei (A>60):
  • – Binding energy per nucleon gradually decreases as A
  • – Example: Uranium-238 [math]{}_{92}^{238}\text{U}[/math] has [math]\frac{E_b}{A} ≈ 7.6MeV[/math]
  • – Larger nuclei are less stable and prone to fission.
  • ⇒   Interpretation:
  • 1. Stability and Energy Release:
  • – Fusion: Light nuclei (e.g., hydrogen) release energy by combining to form heavier nuclei with higher binding energy per nucleon.
  • – Fission: Heavy nuclei (e.g., uranium) release energy by splitting into lighter nuclei with higher binding energy per nucleon.
  • 2. Iron Peak:
  • – The “peak” at Iron-56 indicates the maximum stability of nuclei.
  • – Nuclei lighter than iron can fuse, while nuclei heavier than iron undergo fission to increase binding energy per nucleon.

  • c) Make calculations and estimates involving:

  • I)   Activity of a Sample of Radioactive Material

  • 1. Activity (A):
  • – The number of radioactive decays per second in a sample.
  • – Measured in becquerels (Bq), where 1Bq = 1decay/second.
  • 2. Decay Law:
  • [math]A = A_0 e^{-λt}[/math]
  • – Where:
  • [math]A_o[/math]​: Initial activity (Bq).
  • λ: Decay constant ([math]s^{-1}[/math]).
  • t: Time elapsed (s).
  • 3. Relationship Between Decay Constant and Half-Life:
  • [math]\lambda = \frac{\ln(2)}{T_{1/2}}[/math]
  • – where [math]T_{1/2}[/math]​ is the half-life of the material.
  • ⇒ Example Calculation
  • Problem: A sample of [math]I^{131}[/math] (half-life [math]T_{1/2} = 8 \text{days}[/math]) has an initial activity of 500 Bq. What is the activity after 16 days?
  • 1. Calculate λ:
  • [math]\lambda = \frac{\ln(2)}{T_{1/2}} \\
    \lambda = \frac{\ln(2)}{8 \times 24 \times 3600} \\
    \lambda \approx 1 \times 10^{-6} \text{ s}^{-1}[/math]
  • 2. Calculate A after 16 days (t=16 days):
  • [math]A = A_0 e^{-\lambda t} \\
    A = 500 e^{-(1 \times 10^{-6}) (8 \times 24 \times 3600)}[/math]
  • 3. Simplify:
  • [math]A = (500)e^{-1.386} \\  A = (500)(0.250) \\ A ≈ 125 Bq[/math]
  • Answer: The activity after 16 days is 125 Bq.

  • II)  Absorbed Dose in Gray

  • 1. Absorbed Dose (D):
  • – Energy deposited per unit mass of tissue.
  • – Formula:
  • [math]D = \frac{E}{m}[/math]
  •  Where:
  • E: Energy deposited (in joules).
  • m: Mass of the material (in kilograms).
  • – Measured in grays (Gy), where [math]1 Gy = 1 J/kg[/math].
  • ⇒  Example Calculation
  • Problem: A 2 kg block of tissue absorbs 0.5J of radiation energy. What is the absorbed dose?
  • 1. Use the formula:
  • [math]D = \frac{E}{m} \\
    D = \frac{0.5}{2} \\
    D = 0.25 \text{ Gy}[/math]
  • Answer: The absorbed dose is [math]0.25 \text{Gy}[/math].

  • III)    Effective Dose in Sieverts

  • 1. Effective Dose (H):
  • Accounts for the type of radiation and its biological impact.
  • Formula:
  • [math]H = D ⋅ Q[/math]
  • where:
  • – D: Absorbed dose (in grays).
  • – Q: Radiation quality factor (dimensionless), depends on radiation type.
  • – Q=1 for gamma and beta radiation.
  • – Q=20 for alpha radiation.
  • ⇒  Example Calculation
  • Problem: A tissue receives an absorbed dose of 1 Gy from alpha radiation (Q=20). What is the effective dose?
  • 1. Use the formula:
  • [math]H = D ⋅ Q \\ H = (0.1)(20) \\  H = 2 Sv[/math]
  • Answer: The effective dose is 2.0 Sv.

  • IV)  Energy Changes from Nuclear Transformations

  • 1. Rest Energy (​[math]E_{rest}[/math]):
  • – Energy equivalent to a particle’s mass.
  • – Formula:
  • [math]E_{rest} = mc^2[/math]
  • Where:
  • – m: Mass (in kilograms).
  • – c: Speed of light ([math]3 × 10^8[/math]m/s).
  • 2. Mass Defect and Binding Energy:
  • – The difference in mass between the nucleus and its constituent protons and neutrons ([math]Δm[/math]).
  • – Energy released or absorbed:
  • [math]E_{rest} = mc^2[/math]
  • ⇒  Example Calculation
  • Problem: In a nuclear fission reaction, [math]Δm = 0.001u[/math]. What is the energy released?
  • 1. Convert [math]Δm[/math] to kilograms:
  • [math]\Delta m = 0.001u \times (1.660 \times 10^{-27} \text{ kg/u}) \\
    \Delta m = 1.660 \times 10^{-30} \text{ kg}[/math]
  • 2. Use [math]E = \Delta mc^2[/math]
  • [math]E = \Delta mc^2 \\
    E = (1.660 \times 10^{-30}) (3 \times 10^8)^2 \\
    E = 1.49 \times 10^{-13} \text{ J} [/math]
  • 3. Convert to MeV ([math]1J = 6.242 × 10^12 MeV[/math]):
  • [math]E = 1.49 \times 10^{-13} \text{ J} \\
    E = (1.49 \times 10^{-13}) \times (6.242 \times 10^{12}) \\
    E \approx 0.93 \text{ MeV}[/math]
  • Answer: The energy released is approximately [math]0.93 \text{Mev}[/math].

  • d) Demonstrate and apply knowledge and understanding of the following practical activities (HSW4):

  • I) Studying the Absorption of α-Particles, β-Particles, and γ-Rays by Appropriate Materials

  • 1. Background
  • ⇒ Ionizing Radiation Types:
  • α-Particles (Helium nuclei, [math]{}_{2}^{4}\text{He}[/math] ):
  • – Strongly ionizing but weakly penetrating.
  • – Stopped by a sheet of paper or a few centimeters of air.
  • β-Particles (Electrons or positrons):
  • – Moderately ionizing and penetrating.
  • – Stopped by a few millimeters of aluminum or plastic.
  • γ-Rays (High-energy photons):
  • – Weakly ionizing but highly penetrating.
  • – Partially absorbed by thick lead or several centimeters of concrete.
  • Figure 7 Study of absorbed particles
  • 2. Aim
  • To study the absorption properties of α-, β-, and γ-radiation when passed through different materials.
  • 3. Apparatus
  • Radiation sources for α-, β-, and γ-radiation.
  • Geiger-Müller (GM) tube connected to a counter.
  • Absorbers: paper, aluminum, and lead of varying thickness.
  • Ruler or caliper for measuring distances.
  • 4. Method
  • Setup:
  • – Place the radiation source and GM tube in a fixed alignment.
  • – Ensure no obstacles between the source and detector to establish a baseline count rate.
  • Baseline Measurement:
  • – Record the count rate without any absorber to determine the initial intensity (I0I_0I0​).
  • Introduce Absorbers:
  • – Place different materials (paper, aluminum, lead) of varying thickness between the source and GM tube.
  • – For each absorber, measure and record the count rate (I).
  • Repeat Measurements:
  • – For consistency, repeat each measurement multiple times and take an average.
  • Analyze Results:
  • – Plot graphs of count rate (III) vs. absorber thickness for each type of radiation.
  • Interpret the behavior:
  • α-particles: Rapid drop to zero (stopped by thin material like paper).
  • β-particles: Gradual decrease, requiring denser material (e.g., aluminum).
  • γ-rays: Exponential decrease, requiring very thick materials (e.g., lead).
  • Results
  • – Absorption Properties:
  • α-particles: Stopped by a few centimeters of air or a sheet of paper.
  • β-particles: Require several millimeters of aluminum.
  • γ-rays: Partially absorbed by thick lead; intensity decreases exponentially.

  • II) Determining the Half-Life of an Isotope Such as Protactinium

  • ⇒  Background
  • Half-Life:
  • – Time required for half of the radioactive nuclei in a sample to decay.
  • – Related to the decay constant (λ) by:
  • [math]T_{1/2} = \frac{\ln(2)}{\lambda}[/math]
  • Protactinium-234:
  • – A commonly used isotope in schools, decays via beta emission.
  • – It is safe to use and has a short half-life (~70 seconds).
  • ⇒  Aim
  • To determine the half-life of protactinium-234 using a GM tube and timer.
  • ⇒  Apparatus
  • A protactinium generator (contains uranium salt and an organic solvent).
  • GM tube and counter.
  • Stopwatch or timer.
  • Data recording sheet.
  • Figure 8 Determine radioactivity
  • ⇒  Method
  • 1. Setup:
  • – Shake the protactinium generator to mix the aqueous and organic layers, allowing protactinium-234 to enter the organic layer.
  • – Place the GM tube near the generator to detect beta radiation.
  • 2. Baseline Measurement:
  • – Measure the background radiation count rate ([math]I_{background}[/math]​) without the generator.
  • 3. Start the Timer:
  • – Immediately after shaking, start the stopwatch and record the initial count rate ([math]I_o[/math]​).
  • 4. Measure Decay:
  • – At regular intervals (e.g., every 10 seconds), record the count rate (I) from the generator
  • – Subtract the background count rate from each measurement to find the corrected count rate.
  • 5. Continue Measurements:
  • – Repeat until the count rate becomes close to the background level.
  • 6. Plot Data:
  • – Plot a graph of the corrected count rate (I) vs. time (t).
  • – Alternatively, plot [math]\ln(I)[/math] vs. t for a straight-line relationship:
  • [math]\ln(I) = \ln(I_0) – \lambda t[/math]
  • ⇒  Results
  • Determine the half-life ([math]T_{1/2}[/math]​):
  • From the graph of [math]\ln(I)[/math] t, the slope gives −λ.
  • Calculate ​[math]T_{1/2}[/math] using:
  • [math]T_{1/2} = \frac{\ln(2)}{\lambda}[/math]
For video lectures subscribe to our YouTube channel
Request a lesson
error: Content is protected !!