Nuclear Physics

Atomic Structure Review

The Nuclear Atom

An atom consists of a small, dense, positively charged nucleus surrounded by negatively charged electrons orbiting in shells. The nucleus contains two types of nucleons:

Particle	Symbol	Charge	Mass (u)	Mass (kg)
Proton	$p$	$+1.6 \times 10^{-19}$ C	$1.007276$	$1.673 \times 10^{-27}$
Neutron	$n$	$0$	$1.008665$	$1.675 \times 10^{-27}$
Electron	$e^-$	$-1.6 \times 10^{-19}$ C	$0.000549$	$9.11 \times 10^{-31}$

The atomic mass unit (u) is defined as $\frac{1}{12}$ the mass of a carbon-12 atom:

$1 \mathrm{ u} = 1.66054 \times 10^{-27} \mathrm{ kg}$

The nucleus occupies roughly $10^{-15}$ m of an atom with diameter $10^{-10}$ m. If the atom were the size of a football stadium, the nucleus would be approximately the size of a marble at the centre.

Nuclear Notation

Definition. Nuclear notation represents an atom as $\prescript{A}{}{Z}\mathrm{X}$ where $A$ is the mass number (total nucleons), $Z$ is the atomic number (protons), and X is the chemical symbol.

$\prescript{A}{}{Z}\mathrm{X}$

• $A = Z + N$ where $N$ is the neutron number
• $Z$ determines the element (chemical identity)
• $A$ determines the isotope

Examples:

• $\prescript{235}{}{92}\mathrm{U}$: uranium-235 with 92 protons and 143 neutrons
• $\prescript{1}{}{1}\mathrm{H}$: protium (the most common hydrogen isotope)
• $\prescript{2}{}{1}\mathrm{H}$: deuterium (heavy hydrogen, one proton + one neutron)
• $\prescript{3}{}{1}\mathrm{H}$: tritium (one proton + two neutrons, radioactive)
• $\prescript{12}{}{6}\mathrm{C}$: carbon-12 (the standard for defining the atomic mass unit)

Isotopes, Isobars, and Isotones

Definition. Isotopes are atoms of the same element (same $Z$) but different mass number (different $N$). They have identical chemical properties but different nuclear properties.

Term	Same $Z$	Same $N$	Same $A$
Isotopes	Yes	No	No
Isobars	No	No	Yes
Isotones	No	Yes	No

Examples:

• Isotopes: $\prescript{1}{}{1}\mathrm{H}$, $\prescript{2}{}{1}\mathrm{H}$, $\prescript{3}{}{1}\mathrm{H}$
• Isobars: $\prescript{40}{}{20}\mathrm{Ca}$ and $\prescript{40}{}{18}\mathrm{Ar}$
• Isotones: $\prescript{14}{}{6}\mathrm{C}$ and $\prescript{15}{}{7}\mathrm{N}$ (both have $N = 8$)

info

In DSE exams, isotopes share the same chemical symbol and chemical behaviour. Only nuclear reactions can distinguish between isotopes of the same element.

Nuclear Forces

The nucleus is held together by the strong nuclear force, which is:

• Attractive at distances of $1$ to $3 \mathrm{ fm}$ ($1 \mathrm{ fm} = 10^{-15}$ m)
• Repulsive at distances shorter than about $0.5 \mathrm{ fm}$ (hard core repulsion)
• Independent of charge (acts equally between proton-proton, neutron-neutron, and proton-neutron pairs)
• Much stronger than the electrostatic force at short range, but has a very short range
• Not described by a simple inverse-square law

The competition between the attractive strong nuclear force and the repulsive electrostatic force (between protons) determines nuclear stability. Heavy nuclei with many protons require extra neutrons to provide additional strong force to counteract the increasing electrostatic repulsion.

Worked Example: Nuclear Notation

How many protons, neutrons, and nucleons are in $\prescript{238}{}{92}\mathrm{U}$?

Solution

• Protons: $Z = 92$
• Neutrons: $N = A - Z = 238 - 92 = 146$
• Nucleons: $A = 238$

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Radioactivity

Alpha Decay

Explore the simulation above to develop intuition for this topic.

Nature of Radioactivity

Definition. Radioactivity is the spontaneous disintegration of unstable nuclei with the emission of radiation. It is a random, spontaneous process that depends only on the nuclear structure, not on external conditions such as temperature, pressure, or chemical state.

Radioactivity was discovered by Henri Becquerel in 1896 when he observed that uranium salts could expose photographic plates. Marie and Pierre Curie subsequently isolated polonium and radium.

Types of Radiation

There are three main types of radiation emitted by radioactive nuclei:

Property	Alpha ($\alpha$)	Beta-minus ($\beta^-$)	Beta-plus ($\beta^+$)	Gamma ($\gamma$)
Nature	Helium nucleus $\prescript{4}{}{2}\mathrm{He}^{2+}$	Electron $e^-$	Positron $e^+$	Electromagnetic wave
Charge	$+2e$	$-e$	$+e$	$0$
Mass (u)	$4.0015$	$0.00055$	$0.00055$	$0$
Speed	$\sim 5\%$ of $c$	Up to $99\%$ of $c$	Up to $99\%$ of $c$	$c$ (speed of light)
Ionising power	Very high	Moderate	Moderate	Low
Penetrating power	Very low (stopped by paper or a few cm of air)	Moderate (stopped by a few mm of aluminium)	Moderate (stopped by a few mm of aluminium)	Very high (requires thick lead/concrete)
Range in air	$\sim 5 \mathrm{ cm}$	$\sim 1 \mathrm{ m}$	$\sim 1 \mathrm{ m}$	Infinite (intensity decreases with $1/r^2$)
Deflection in E/B field	Deflected towards negative plate	Deflected towards positive plate	Deflected towards negative plate	Not deflected
Energy spectrum	Discrete (monoenergetic)	Continuous (shared with antineutrino)	Continuous (shared with neutrino)	Discrete (line spectrum)

Alpha Decay

In alpha decay, the nucleus emits an alpha particle ($\prescript{4}{}{2}\mathrm{He}$), reducing both $A$ by $4$ and $Z$ by $2$:

$\prescript{A}{}{Z}\mathrm{X} \to \prescript{A-4}{}{Z-2}\mathrm{Y} + \prescript{4}{}{2}\mathrm{He}$

The daughter nucleus shifts two places to the left in the periodic table.

Example (radium-226 decay):

$\prescript{226}{}{88}\mathrm{Ra} \to \prescript{222}{}{86}\mathrm{Rn} + \prescript{4}{}{2}\mathrm{He}$

Alpha particles are emitted with a single characteristic energy (discrete spectrum) because the transition is between two well-defined nuclear energy levels. Alpha decay occurs primarily in heavy nuclei ($A \gt 150$) where the strong nuclear force can no longer overcome the electrostatic repulsion.

Beta-Minus Decay

In beta-minus decay, a neutron inside the nucleus converts into a proton, emitting an electron and an antineutrino:

$n \to p + e^- + \bar{\nu}_e$

The nuclear equation is:

$\prescript{A}{}{Z}\mathrm{X} \to \prescript{A}{}{Z+1}\mathrm{Y} + e^- + \bar{\nu}_e$

The daughter nucleus shifts one place to the right in the periodic table. The mass number $A$ does not change because a neutron is replaced by a proton.

Beta-minus decay occurs when the nucleus has an excess of neutrons (neutron-rich nuclei). The energy is shared between the beta particle and the antineutrino, producing a continuous energy spectrum for the beta particle.

Why is the antineutrino necessary? Without the antineutrino, both energy and momentum conservation would be violated. The continuous energy spectrum of beta particles (unlike the discrete alpha spectrum) was evidence that a third particle carries away some energy.

Beta-Plus (Positron) Decay

In beta-plus decay, a proton inside the nucleus converts into a neutron, emitting a positron and a neutrino:

$p \to n + e^+ + \nu_e$

The nuclear equation is:

$\prescript{A}{}{Z}\mathrm{X} \to \prescript{A}{}{Z-1}\mathrm{Y} + e^+ + \nu_e$

Beta-plus decay occurs in proton-rich nuclei. The daughter nucleus shifts one place to the left in the periodic table.

Condition for beta-plus decay: The parent nucleus must have enough mass-energy to create the positron. Specifically:

$m(\mathrm{parent}) \gt m(\mathrm{daughter}) + 2m_e$

The extra $m_e$ is required because the daughter has one fewer electron, so one orbital electron must be emitted as well. If this condition is not met, electron capture may occur instead:

$\prescript{A}{}{Z}\mathrm{X} + e^- \to \prescript{A}{}{Z-1}\mathrm{Y} + \nu_e$

Gamma Radiation

Gamma rays are high-energy photons emitted when a nucleus transitions from an excited state to a lower energy state:

$\prescript{A}{}{Z}\mathrm{X}^* \to \prescript{A}{}{Z}\mathrm{X} + \gamma$

The asterisk denotes an excited nuclear state. Gamma emission does not change $A$ or $Z$. It typically follows alpha or beta decay when the daughter nucleus is left in an excited state.

Gamma rays have a discrete line spectrum because the nuclear energy levels are quantised. They are the most penetrating form of radiation but the least ionising.

Conservation Laws in Nuclear Decay

Every nuclear decay must satisfy:

1. Conservation of nucleon number ($A$): total $A$ is conserved
2. Conservation of proton number ($Z$): total $Z$ is conserved
3. Conservation of charge: total charge is conserved
4. Conservation of mass-energy: total mass-energy is conserved (Q-value)
5. Conservation of momentum: linear momentum is conserved
6. Conservation of lepton number: total lepton number is conserved

warning

warning equations. While DSE exams sometimes omit them for simplicity, always check whether the question requires them. Also, ensure $A$ and $Z$ balance on both sides of every decay equation.

Worked Example: Balancing Decay Equations

Complete the following decay equation: $\prescript{214}{}{84}\mathrm{Po} \to \prescript{210}{}{82}\mathrm{Pb} + \ ?$

Solution

Check the nucleon number: $214 - 210 = 4$

Check the proton number: $84 - 82 = 2$

The missing particle has $A = 4$, $Z = 2$, which is an alpha particle: $\prescript{4}{}{2}\mathrm{He}$

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Radioactive Decay

Random and Spontaneous Nature

Radioactive decay is random: it is impossible to predict which specific nucleus will decay next or exactly when a particular nucleus will decay. It is spontaneous: the decay is not affected by external conditions such as temperature, pressure, chemical bonding, or physical state.

These properties are confirmed experimentally by:

• The fluctuation in count rate observed with a Geiger-Muller tube (statistical fluctuations)
• The identical decay rate of a radioactive compound regardless of its chemical form

Activity and Decay Constant

Definition. The activity $A$ of a radioactive source is the number of decays per unit time.

$A = -\frac{dN}{dt} = \lambda N$

Where:

• $N$ = number of undecayed nuclei at time $t$
• $\lambda$ = decay constant (probability of decay per nucleus per unit time)
• The SI unit of activity is the becquerel (Bq), where $1 \mathrm{ Bq} = 1 \mathrm{ decay/s}$

The decay constant $\lambda$ is characteristic of a particular isotope. A large $\lambda$ means the isotope decays quickly (short-lived); a small $\lambda$ means it decays slowly (long-lived).

Exponential Decay Law

Starting from $A = \lambda N$ and solving the differential equation:

$\frac{dN}{dt} = -\lambda N$

We obtain:

$N = N_0 e^{-\lambda t}$

Where $N_0$ is the initial number of undecayed nuclei at $t = 0$.

Since activity is proportional to the number of undecayed nuclei ($A = \lambda N$), activity also follows exponential decay:

$A = A_0 e^{-\lambda t}$

Where $A_0 = \lambda N_0$ is the initial activity.

Half-Life

Definition. The half-life $t_{1/2}$ is the time taken for half of the radioactive nuclei in a sample to decay.

Setting $N = \frac{N_0}{2}$ in the decay law:

$\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}}$

$\frac{1}{2} = e^{-\lambda t_{1/2}}$

$\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2}$

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

Conversely:

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

After $n$ half-lives, the fraction remaining is:

$\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \frac{1}{2^n}$

Number of half-lives ($n$)	Fraction remaining	Percentage remaining
$0$	$1$	$100\%$
$1$	$1/2$	$50\%$
$2$	$1/4$	$25\%$
$3$	$1/8$	$12.5\%$
$4$	$1/16$	$6.25\%$
$5$	$1/32$	$3.125\%$

Determining Half-Life from a Graph

Given an exponential decay curve of activity (or count rate) versus time:

1. Find the initial activity $A_0$
2. Find the time at which the activity drops to $\frac{A_0}{2}$ -- this is $t_{1/2}$
3. Verify by checking that the activity at $2t_{1/2}$ is $\frac{A_0}{4}$

For a logarithmic plot ($\ln A$ vs $t$):

$\ln A = \ln A_0 - \lambda t$

This is a straight line with:

• Slope $= -\lambda$
• y-intercept $= \ln A_0$

The half-life is then:

$t_{1/2} = \frac{\ln 2}{\lvert\mathrm{slope}\rvert}$

info

info plot, remember the slope is negative: $\lvert\mathrm{slope}\rvert = \lambda$. Always check the axes carefully -- count rate is proportional to activity but is lower due to detector efficiency.

Relationship Between Activity and Mass

For a sample of mass $m$ of an isotope with molar mass $M$:

$N = \frac{m}{M} N_A$

Where $N_A = 6.02 \times 10^{23} \mathrm{ mol}^{-1}$ is the Avogadro constant.

The activity is then:

$A = \lambda N = \frac{\lambda m N_A}{M} = \frac{m N_A \ln 2}{M t_{1/2}}$

warning

warning detected by a particular instrument). The count rate is always less than or equal to the activity because:

• The detector only captures a fraction of emitted radiation (solid angle)
• Not all radiation reaches the detector (absorption by air, source holder)
• The detector has less than 100% efficiency

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Nuclear Reactions

Mass-Energy Equivalence

Einstein's mass-energy equivalence is the foundational principle behind all nuclear energy calculations:

$E = mc^2$

Where:

• $E$ = energy equivalent (J)
• $m$ = mass (kg)
• $c = 3.0 \times 10^8 \mathrm{ m/s}$ (speed of light)

A small amount of mass corresponds to a very large amount of energy. For nuclear physics calculations, it is often convenient to use the conversion:

$1 \mathrm{ u} \times c^2 = 931.5 \mathrm{ MeV}$

or equivalently:

$1 \mathrm{ MeV}/c^2 = 1.783 \times 10^{-30} \mathrm{ kg}$

Mass Defect and Binding Energy

Definition. The mass defect $\Delta m$ of a nucleus is the difference between the total mass of its constituent nucleons (when separated) and the actual mass of the nucleus:

$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}$

Where:

• $Z$ = number of protons
• $N$ = number of neutrons
• $m_p$ = mass of a proton
• $m_n$ = mass of a neutron
• $m_{\mathrm{nucleus}}$ = actual mass of the nucleus

The mass defect is always positive for stable nuclei. The "missing" mass has been converted into binding energy that holds the nucleus together.

Definition. The binding energy $BE$ of a nucleus is the energy equivalent of its mass defect:

$BE = \Delta m \cdot c^2$

This is the minimum energy required to completely separate the nucleus into its individual protons and neutrons.

Definition. The binding energy per nucleon is the binding energy divided by the mass number:

$\frac{BE}{A} = \frac{\Delta m \cdot c^2}{A}$

This is a measure of nuclear stability -- the higher the binding energy per nucleon, the more stable the nucleus.

Binding Energy per Nucleon Curve

The binding energy per nucleon curve is one of the most important graphs in nuclear physics:

Region	Mass number range	Binding energy per nucleon	Characteristics
Very light nuclei	$A \lt 20$	Rising sharply	Fusion releases energy
Peak region (Fe-56)	$A \approx 50-60$	Maximum ($\sim 8.8$ MeV)	Most stable nuclei
Medium nuclei	$20 \lt A \lt 60$	Relatively flat	Stable
Heavy nuclei	$A \gt 60$	Gradually decreasing	Fission releases energy
Very heavy nuclei	$A \gt 200$	$\sim 7.5$ MeV	Unstable, can undergo fission

Key points:

• Iron-56 ($\prescript{56}{}{26}\mathrm{Fe}$) has the highest binding energy per nucleon and is the most stable nucleus
• Energy is released when light nuclei fuse (move towards the peak from the left)
• Energy is released when heavy nuclei fission (move towards the peak from the right)
• The curve explains why both fusion and fission can release energy

Nuclear Fission

Definition. Nuclear fission is the splitting of a heavy nucleus into two (or occasionally three) lighter nuclei, accompanied by the release of energy and typically two or three neutrons.

The most studied fission reaction is uranium-235:

$\prescript{1}{}{0}\mathrm{n} + \prescript{235}{}{92}\mathrm{U} \to \prescript{236}{}{92}\mathrm{U}^* \to \prescript{141}{}{56}\mathrm{Ba} + \prescript{92}{}{36}\mathrm{Kr} + 3\prescript{1}{}{0}\mathrm{n} + \mathrm{energy}$

The released neutrons can induce further fission reactions, creating a chain reaction.

Critical mass is the minimum mass of fissile material required to sustain a chain reaction. If the mass is subcritical, too many neutrons escape without causing further fission. If the mass is supercritical, the reaction rate increases exponentially.

Components of a Nuclear Reactor

Component	Function	Material examples
Fuel rods	Contain fissile material	U-235, Pu-239
Moderator	Slow down fast neutrons to thermal energies for efficient fission	Water, heavy water, graphite
Control rods	Absorb excess neutrons to control the reaction rate	Boron, cadmium, hafnium
Coolant	Remove heat from the reactor core	Water, liquid sodium, CO$_2$
Containment	Prevent radiation leaks to the environment	Thick concrete and steel
Shielding	Absorb gamma rays and neutrons	Lead, concrete, water

Moderator: Fast neutrons from fission have energies of about $2 \mathrm{ MeV}$. U-235 fission is much more probable with thermal (slow) neutrons ($\sim 0.025 \mathrm{ eV}$). The moderator slows neutrons through elastic collisions. A good moderator has a small mass number (for efficient energy transfer in elastic collisions) and a low neutron absorption cross-section.

Control rods: These are inserted or withdrawn to regulate the reaction rate. Inserting control rods absorbs more neutrons, reducing the reaction rate. Withdrawing them allows more neutrons to cause fission, increasing the rate. In an emergency (SCRAM), control rods are fully inserted to shut down the reactor.

Types of Nuclear Power Reactors

Pressurised Water Reactor (PWR):

• Water acts as both coolant and moderator
• Primary coolant loop is kept at high pressure ($\sim 155$ bar) to prevent boiling
• Heat is transferred to a secondary loop via a heat exchanger to produce steam
• Most common reactor type worldwide
• Negative temperature coefficient provides inherent safety

Boiling Water Reactor (BWR):

• Water boils directly in the reactor core to produce steam
• Steam drives the turbine directly (no secondary loop)
• Simpler design but radioactive steam passes through the turbine
• Lower operating pressure than PWR

Feature	PWR	BWR
Coolant	Pressurised water (does not boil)	Boiling water
Steam generation	Secondary loop (heat exchanger)	Direct in reactor core
Moderator	Water (primary loop)	Water
Pressure	$\sim 155$ bar	$\sim 70$ bar
Fuel enrichment	$3$-$5\%$ U-235	$2$-$4\%$ U-235
Radioactive steam	No (secondary loop is clean)	Yes (steam is radioactive)

Nuclear Fusion

Definition. Nuclear fusion is the combining of two light nuclei to form a heavier nucleus, releasing energy in the process.

Fusion releases energy because the product nucleus has a higher binding energy per nucleon than the reactants (moving towards the peak of the binding energy curve).

Example fusion reactions:

$\prescript{2}{}{1}\mathrm{H} + \prescript{2}{}{1}\mathrm{H} \to \prescript{3}{}{2}\mathrm{He} + \prescript{1}{}{0}\mathrm{n} + 3.27 \mathrm{ MeV}$

$\prescript{2}{}{1}\mathrm{H} + \prescript{3}{}{1}\mathrm{H} \to \prescript{4}{}{2}\mathrm{He} + \prescript{1}{}{0}\mathrm{n} + 17.6 \mathrm{ MeV}$

Conditions for Fusion

For fusion to occur, nuclei must overcome the electrostatic repulsion between them. This requires:

1. High temperature ($\sim 10^7$ to $10^8$ K): nuclei must have sufficient kinetic energy to overcome the Coulomb barrier
2. High density: increases the collision rate between nuclei
3. Confinement time: nuclei must be held together long enough for fusion to occur

These three conditions are described by the Lawson criterion.

Fusion in Stars

Stars are powered by fusion. The main processes are:

Proton-proton (pp) chain (dominant in stars like the Sun):

Step 1: $\prescript{1}{}{1}\mathrm{H} + \prescript{1}{}{1}\mathrm{H} \to \prescript{2}{}{1}\mathrm{H} + e^+ + \nu_e + 0.42 \mathrm{ MeV}$

Step 2: $\prescript{2}{}{1}\mathrm{H} + \prescript{1}{}{1}\mathrm{H} \to \prescript{3}{}{2}\mathrm{He} + \gamma + 5.49 \mathrm{ MeV}$

Step 3: $\prescript{3}{}{2}\mathrm{He} + \prescript{3}{}{2}\mathrm{He} \to \prescript{4}{}{2}\mathrm{He} + 2\prescript{1}{}{1}\mathrm{H} + 12.86 \mathrm{ MeV}$

Net: $4\prescript{1}{}{1}\mathrm{H} \to \prescript{4}{}{2}\mathrm{He} + 2e^+ + 2\nu_e + 2\gamma + 26.7 \mathrm{ MeV}$

CNO cycle (dominant in stars more massive than the Sun):

Uses carbon, nitrogen, and oxygen as catalysts. The net result is the same: four protons fuse to form a helium-4 nucleus with the release of about $26.7 \mathrm{ MeV}$. The CNO cycle has a much stronger temperature dependence than the pp chain, making it dominant at higher temperatures.

Q-Value of Nuclear Reactions

Definition. The Q-value of a nuclear reaction is the energy released (or absorbed) in the reaction, calculated from the mass difference between reactants and products:

$Q = (m_{\mathrm{reactants}} - m_{\mathrm{products}}) \times c^2$

• $Q \gt 0$: exothermic reaction (energy released, e.g., fission, fusion)
• $Q \lt 0$: endothermic reaction (energy absorbed, threshold energy required)

For the D-T fusion reaction:

$Q = [m(\prescript{2}{}{1}\mathrm{H}) + m(\prescript{3}{}{1}\mathrm{H}) - m(\prescript{4}{}{2}\mathrm{He}) - m(\prescript{1}{}{0}\mathrm{n})]c^2 = 17.6 \mathrm{ MeV}$

info

In DSE calculations, always convert masses to the same units (preferably u) before computing the Q-value. Use $1 \mathrm{ u} = 931.5 \mathrm{ MeV}/c^2$ for the energy conversion. Remember that the Q-value is shared among all products as kinetic energy (and possibly photons).

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Detection and Measurement

Geiger-Muller Tube

The Geiger-Muller (GM) tube is the most commonly used radiation detector in school laboratories.

Construction and operation:

1. A thin mica window at one end allows radiation to enter
2. The tube contains a low-pressure gas (typically argon with a small amount of halogen quenching gas)
3. A central anode wire is at high positive potential ($\sim 400$-$900$ V); the cylindrical cathode is at ground potential
4. When radiation enters the tube, it ionises gas atoms, producing ion pairs
5. The electrons are accelerated towards the anode and ionise more gas atoms in an avalanche (Townsend avalanche)
6. Each avalanche produces a detectable pulse of current
7. The pulse is counted by an electronic counter

Dead time: After each detection event, the GM tube requires a brief recovery period ($\sim 100$ to $300 \ \mu\mathrm{s}$) during which it cannot detect new events. This is called the dead time. At high count rates, some events are missed, leading to an undercount.

Quenching: Without quenching, the positive ions would reach the cathode and release secondary electrons, causing multiple pulses from a single radiation event. Quenching is achieved by:

• Adding a small amount of halogen gas (self-quenching tube) that absorbs the energy of positive ions
• Using an external quenching circuit

Limitations of the GM tube:

• Cannot distinguish between different types of radiation
• Cannot measure the energy of radiation
• Has a dead time that limits the maximum count rate
• Cannot detect very low-energy radiation (absorbed by the window)

Photographic Film

Photographic film darkens when exposed to ionising radiation. The degree of darkening depends on the total dose received.

• Used in film badges worn by radiation workers to monitor cumulative exposure
• Can provide a permanent record of radiation exposure
• Different filters (e.g., aluminium, lead) over different sections allow estimation of the type and energy of radiation
• Simple, inexpensive, and requires no power supply
• Cannot provide real-time readings; must be developed in a laboratory

Scintillation Counter

A scintillation counter uses a scintillating material that emits flashes of light (scintillations) when ionising radiation passes through it.

Operation:

1. Radiation strikes a scintillator (e.g., sodium iodide doped with thallium, NaI(Tl))
2. The scintillator produces a flash of light
3. The light is detected by a photomultiplier tube (PMT)
4. The PMT converts the light into an electrical signal and amplifies it enormously

Advantages over GM tube:

• Can measure the energy of radiation (pulse height analysis)
• Faster response time (shorter dead time)
• Higher detection efficiency for certain types of radiation
• Can distinguish between different types of radiation based on pulse characteristics

Cloud Chamber

A cloud chamber makes the paths of ionising radiation visible by creating a supersaturated vapour.

Operation:

1. A chamber contains alcohol (or water) vapour near its condensation point
2. The chamber is cooled rapidly (expansion-type) or has a cold plate (diffusion-type)
3. When ionising radiation passes through the chamber, it ionises air molecules along its path
4. The ions act as condensation nuclei, and tiny liquid droplets form along the track
5. The tracks are illuminated and can be photographed or observed directly

Track characteristics:

Radiation	Track appearance	Length
Alpha	Thick, straight, dense track	Short ($\sim 5$ cm)
Beta	Thin, winding track (easily deflected)	Long ($\sim 1$ m)
Gamma	Very faint, short, scattered tracks (pair production, Compton)	Very short

Semiconductor Detector

Semiconductor detectors use the principle that ionising radiation creates electron-hole pairs in a semiconductor material (e.g., silicon or germanium).

• Each ionising event creates many electron-hole pairs, proportional to the energy deposited
• Very good energy resolution
• Fast response
• Compact size
• Often requires cooling (especially germanium detectors) to reduce thermal noise

Comparison of Detectors

Detector	Radiation types detected	Energy measurement	Advantages	Disadvantages
GM tube	Alpha, beta, gamma	No	Simple, portable, cheap	No energy resolution, dead time
Photographic film	Alpha, beta, gamma, X-rays	No	Permanent record, no power needed	No real-time reading, slow
Scintillation counter	Alpha, beta, gamma	Yes	Good energy resolution, fast	Expensive, requires PMT
Cloud chamber	Alpha, beta, gamma	Limited	Visualises tracks	Bulky, requires careful setup
Semiconductor	Alpha, beta, gamma, X-rays	Yes (excellent)	Best energy resolution, compact	Expensive, often requires cooling

info

In DSE exams, the GM tube is the most important detector. Know its construction, operation principle, dead time, and limitations. Be prepared to explain why a GM tube cannot distinguish between alpha and beta radiation.

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Biological Effects of Radiation

Ionising Radiation and Damage

Ionising radiation carries enough energy to remove electrons from atoms, creating ions. This ionisation can damage biological tissue through several mechanisms:

1. Direct ionisation: Radiation directly ionises DNA molecules, causing strand breaks
2. Indirect ionisation: Radiation ionises water molecules (the most abundant molecule in the body), producing reactive free radicals ($\mathrm{OH}^*$, $\mathrm{H}^*$) that attack DNA

Types of DNA damage:

• Single-strand breaks: usually repairable
• Double-strand breaks: more serious, may lead to cell death or mutations
• Base damage: may cause mispairing during replication
• Chromosome aberrations: visible under a microscope

Dose Units

Quantity	Unit	Definition
Absorbed dose	Gray (Gy)	Energy absorbed per unit mass: $1 \mathrm{ Gy} = 1 \mathrm{ J/kg}$
Equivalent dose (dose equivalent)	Sievert (Sv)	Absorbed dose weighted by radiation type: $H = D \times w_R$
Effective dose	Sievert (Sv)	Equivalent dose weighted by tissue sensitivity: $E = \sum H_T w_T$
Activity	Becquerel (Bq)	One decay per second: $1 \mathrm{ Bq} = 1 \mathrm{ s}^{-1}$

Radiation weighting factors $w_R$:

Radiation type	$w_R$
X-rays, gamma rays	$1$
Beta particles	$1$
Thermal neutrons	$1$
Fast neutrons, protons	$2$-$20$ (depends on energy)
Alpha particles	$20$

Alpha particles have a high $w_R$ because of their high ionising power, which causes concentrated damage along a short track.

Common dose conversions:

$1 \mathrm{ Sv} = 1000 \mathrm{ mSv}$

$1 \mathrm{ mSv} = 1000 \ \mu\mathrm{Sv}$

Exposure Limits

Category	Annual limit (typical)
General public	$1 \mathrm{ mSv}$
Radiation workers (occupational)	$20 \mathrm{ mSv}$ averaged over 5 years ($50 \mathrm{ mSv}$ in any single year)
Pregnant radiation workers	$1 \mathrm{ mSv}$ to the foetus

Typical radiation doses:

Source	Typical dose
Chest X-ray	$0.02$-$0.1 \mathrm{ mSv}$
Dental X-ray	$0.005 \mathrm{ mSv}$
CT scan (abdomen)	$8$-$10 \mathrm{ mSv}$
Background radiation (per year)	$2.4$-$3.0 \mathrm{ mSv}$
Flight at 35,000 ft (per hour)	$0.003$-$0.005 \mathrm{ mSv}$
Mammogram	$0.4 \mathrm{ mSv}$

ALARA Principle

Definition. The ALARA principle states that radiation exposure should be kept As Low As Reasonably Achievable. This means:

• Time: Minimise time spent near radioactive sources
• Distance: Maximise distance from sources (intensity falls off as $1/r^2$ for a point source)
• Shielding: Use appropriate shielding between the source and the person

Shielding

The choice of shielding material depends on the type of radiation:

Radiation	Shielding material	Reason
Alpha	Paper, human skin	Alpha particles have low penetration; stopped by a few cm of air
Beta	Aluminium ($\sim 1$-$5$ mm) or perspex	Beta particles are stopped by low-Z materials; high-Z materials can produce bremsstrahlung X-rays
Gamma	Lead ($\sim$ cm) or thick concrete	High penetration requires dense, high-Z materials; intensity reduced exponentially

Half-value thickness (HVT): The thickness of material that reduces the intensity of gamma radiation to half its original value:

$I = I_0 e^{-\mu x}$

$\mathrm{HVT} = \frac{\ln 2}{\mu}$

Where $\mu$ is the linear attenuation coefficient.

warning

Never use lead shielding for beta radiation. High-Z materials like lead produce bremsstrahlung (breaking radiation) when beta particles decelerate rapidly, creating X-rays that are more penetrating than the original beta particles. Use aluminium or perspex for beta shielding instead.

---

Applications

Radiocarbon Dating

Carbon-14 dating is used to determine the age of organic materials up to about $50,000$ years.

Principle:

• Carbon-14 ($\prescript{14}{}{6}\mathrm{C}$) is produced in the upper atmosphere by cosmic ray neutrons interacting with nitrogen-14:

  $\prescript{1}{}{0}\mathrm{n} + \prescript{14}{}{7}\mathrm{N} \to \prescript{14}{}{6}\mathrm{C} + \prescript{1}{}{1}\mathrm{H}$

• C-14 is radioactive and undergoes beta-minus decay with a half-life of $5730$ years:

  $\prescript{14}{}{6}\mathrm{C} \to \prescript{14}{}{7}\mathrm{N} + e^- + \bar{\nu}_e$

• Living organisms continuously exchange carbon with the environment, maintaining a constant ratio of C-14 to C-12 ($\sim 1.3 \times 10^{-12}$)

• When an organism dies, it stops exchanging carbon. The C-14 decays while C-12 remains constant

• The ratio of C-14 to C-12 decreases over time, allowing the age to be calculated

Age calculation:

$N = N_0 e^{-\lambda t}$

$\frac{N}{N_0} = e^{-\lambda t}$

$t = \frac{1}{\lambda} \ln\left(\frac{N_0}{N}\right) = \frac{t_{1/2}}{\ln 2} \ln\left(\frac{N_0}{N}\right)$

Limitations:

• Maximum useful age is about $50,000$ years (after about 10 half-lives, too little C-14 remains)
• Requires calibration with other dating methods (tree rings, ice cores) because atmospheric C-14 concentration has varied over time
• Contamination by modern or old carbon can affect results
• Only works for organic materials (formerly living things)

Medical Isotopes

Isotope	Half-life	Radiation emitted	Application
I-131	$8.02$ days	Beta-minus, gamma	Treatment of thyroid cancer and hyperthyroidism
Tc-99m	$6.01$ hours	Gamma ($140$ keV)	Diagnostic imaging (most widely used medical isotope)
Co-60	$5.27$ years	Beta-minus, gamma	Radiotherapy (external beam), sterilisation of equipment
P-32	$14.3$ days	Beta-minus	Treatment of certain blood disorders
Sr-90	$28.8$ years	Beta-minus	Treatment of eye diseases, superficial radiotherapy
F-18	$110$ min	Beta-plus	PET scans (FDG: fluorodeoxyglucose)

Technetium-99m is the most widely used diagnostic radioisotope because:

• Its half-life ($6.01$ hours) is short enough to minimise patient dose but long enough for imaging procedures
• It emits a single, well-defined gamma ray at $140$ keV, ideal for gamma camera detection
• It can be attached to various pharmaceutical compounds to target specific organs
• It is produced from a Mo-99/Tc-99m generator, making it readily available in hospitals

Iodine-131 is used therapeutically because:

• The thyroid gland selectively absorbs iodine, so I-131 concentrates in thyroid tissue
• Beta particles deliver a localised radiation dose to thyroid cells (treating cancer)
• Gamma rays allow imaging of the thyroid for diagnostic purposes

Nuclear Power Plants

Nuclear power plants generate electricity using the heat from controlled nuclear fission. The basic process:

1. Fission in the reactor core produces heat
2. A coolant (water, liquid sodium, CO$_2$) transfers heat from the core
3. The heat is used to produce steam (either directly or via a heat exchanger)
4. Steam drives a turbine connected to a generator
5. The steam is condensed and recycled

Advantages of nuclear power:

• Very high energy density: $1 \mathrm{ kg}$ of U-235 produces as much energy as about $2.7$ million kg of coal
• No greenhouse gas emissions during operation (CO$_2$-free electricity generation)
• Reliable baseload power (not dependent on weather)
• Relatively small fuel volume compared to fossil fuels

Disadvantages of nuclear power:

• Production of long-lived radioactive waste (some isotopes have half-lives of thousands of years)
• Risk of catastrophic accidents (Chernobyl 1986, Fukushima 2011)
• High initial construction costs and long construction times
• Potential for nuclear weapons proliferation (enriched uranium and plutonium)
• Decommissioning costs and challenges

Smoke Detectors

Domestic smoke detectors commonly use a small amount of americium-241 (an alpha emitter):

• Am-241 emits alpha particles that ionise the air in a small chamber
• The ionised air conducts a small current between two electrodes
• When smoke particles enter the chamber, they attach to ions, reducing the current
• The drop in current triggers the alarm

Am-241 has a half-life of $432$ years, so the source lasts for the lifetime of the detector. The alpha particles cannot penetrate the detector casing, so there is no external radiation hazard under normal operation.

---

DSE Exam Focus

Common Question Types

Type 1: Decay Equations and Conservation Laws

Given a decay equation, identify the missing particle. Check:

• Does $A$ balance? (total mass number conserved)
• Does $Z$ balance? (total proton number conserved)
• Is the emitted particle consistent with the decay type?

Type 2: Half-Life Calculations

• Given initial and final activity (or count rate), find the time elapsed
• Given half-life and initial quantity, find the quantity after a given time
• Determine half-life from a graph (linear or log-linear)

Type 3: Binding Energy and Mass Defect

• Calculate mass defect given nuclear masses
• Calculate binding energy per nucleon
• Determine whether fission or fusion is energetically favourable
• Calculate Q-value of a nuclear reaction

Type 4: Activity and Count Rate

• Calculate activity from the number of nuclei and decay constant
• Relate count rate to activity (accounting for efficiency)
• Determine the number of nuclei from activity and half-life

Type 5: Radiocarbon Dating

• Calculate the age of a sample given the current activity and the expected initial activity
• Understand the assumptions and limitations of carbon dating

Type 6: Nuclear Power and Radiation Safety

• Explain the role of moderator, control rods, and coolant
• Compare fission and fusion
• Calculate shielding thickness using half-value thickness

Graph Interpretation Skills

1. Exponential decay graph ($N$ or $A$ vs $t$): Verify that it is exponential by checking that halving the activity always takes the same time. Read half-life directly from the graph.

2. Log-linear graph ($\ln A$ vs $t$): Should be a straight line with negative slope $= -\lambda$. Use the slope to find $\lambda$ and then $t_{1/2}$.

3. Binding energy per nucleon curve: Identify the region of maximum stability (around Fe-56). Determine whether fusion or fission is energetically favourable for a given nucleus.

4. Alpha energy spectrum: Discrete lines at specific energies.

5. Beta energy spectrum: Continuous distribution from zero to a maximum energy $E_{\max}$. The "missing" energy is carried by the neutrino/antineutrino.

Experimental Skills

• Use a GM tube and counter to measure count rate
• Determine half-life from experimental data
• Plot a log-linear graph and extract the decay constant
• Account for background radiation by subtracting the background count rate
• Understand sources of error: statistical fluctuations, dead time, geometry, absorption

info

info calculations and explanations. In Paper 2, it appears as multiple-choice questions testing concepts, definitions, and quick calculations. Practise balancing decay equations and calculating binding energy -- these are high-frequency topics.

Key Formulae Summary

Quantity	Formula
Activity	$A = \lambda N$
Decay law	$N = N_0 e^{-\lambda t}$
Activity decay	$A = A_0 e^{-\lambda t}$
Half-life	$t_{1/2} = \frac{\ln 2}{\lambda}$
Decay constant	$\lambda = \frac{\ln 2}{t_{1/2}}$
Number of nuclei	$N = \frac{m}{M} N_A$
Mass-energy equivalence	$E = mc^2$
Mass defect	$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}$
Binding energy	$BE = \Delta m \cdot c^2$
Q-value	$Q = (m_{\mathrm{reactants}} - m_{\mathrm{products}})c^2$
Energy conversion	$1 \mathrm{ u} = 931.5 \mathrm{ MeV}$
Radiation intensity	$I = I_0 e^{-\mu x}$
Half-value thickness	$\mathrm{HVT} = \frac{\ln 2}{\mu}$

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Worked Examples

Worked Example 1: Alpha Decay Equation

Radon-222 decays by alpha emission. Write the complete decay equation and calculate the energy released if the mass of Rn-222 is $221.970$ u, the mass of Po-218 is $217.963$ u, and the mass of He-4 is $4.003$ u.

Solution

Decay equation:

$\prescript{222}{}{86}\mathrm{Rn} \to \prescript{218}{}{84}\mathrm{Po} + \prescript{4}{}{2}\mathrm{He}$

Energy released (Q-value):

$Q = (m_{\mathrm{parent}} - m_{\mathrm{daughter}} - m_{\alpha}) \times 931.5 \mathrm{ MeV}$

$Q = (221.970 - 217.963 - 4.003) \times 931.5$

$Q = 0.004 \times 931.5 = 3.73 \mathrm{ MeV}$

This energy is shared as kinetic energy between the alpha particle and the polonium-218 daughter nucleus, with most going to the alpha particle (due to conservation of momentum and the lighter mass of the alpha particle).

Worked Example 2: Half-Life Calculation

A radioactive isotope has an initial activity of $800$ Bq. After $30$ minutes, the activity has fallen to $100$ Bq. Calculate the half-life of the isotope.

Solution

Method 1: Using the fraction remaining

The activity has decreased from $800$ Bq to $100$ Bq, so the fraction remaining is:

$\frac{A}{A_0} = \frac{100}{800} = \frac{1}{8} = \frac{1}{2^3}$

This corresponds to $3$ half-lives. Therefore:

$3 t_{1/2} = 30 \mathrm{ min}$

$t_{1/2} = 10 \mathrm{ min}$

Method 2: Using the decay law

$A = A_0 e^{-\lambda t}$

$100 = 800 e^{-\lambda \times 1800}$

$\frac{1}{8} = e^{-1800\lambda}$

$\ln\left(\frac{1}{8}\right) = -1800\lambda$

$-2.079 = -1800\lambda$

$\lambda = 1.155 \times 10^{-3} \mathrm{ s}^{-1}$

$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.155 \times 10^{-3}} = 600 \mathrm{ s} = 10 \mathrm{ min}$

Worked Example 3: Binding Energy per Nucleon

Calculate the binding energy per nucleon of helium-4 ($\prescript{4}{}{2}\mathrm{He}$).

Given:

• Mass of He-4 nucleus $= 4.001506$ u
• Mass of proton $= 1.007276$ u
• Mass of neutron $= 1.008665$ u

Solution

Step 1: Calculate the mass defect

$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}$

$\Delta m = 2(1.007276) + 2(1.008665) - 4.001506$

$\Delta m = 2.014552 + 2.017330 - 4.001506$

$\Delta m = 4.031882 - 4.001506 = 0.030376 \mathrm{ u}$

Step 2: Calculate the binding energy

$BE = \Delta m \times 931.5 \mathrm{ MeV} = 0.030376 \times 931.5 = 28.30 \mathrm{ MeV}$

Step 3: Calculate the binding energy per nucleon

$\frac{BE}{A} = \frac{28.30}{4} = 7.07 \mathrm{ MeV/nucleon}$

Worked Example 4: Radiocarbon Dating

A piece of ancient wood has a carbon-14 activity of $1.5$ Bq per gram of carbon. Living wood has a carbon-14 activity of $15.0$ Bq per gram of carbon. Calculate the age of the ancient wood. (Take $t_{1/2}$ of C-14 $= 5730$ years.)

Solution

Using the decay law:

$A = A_0 e^{-\lambda t}$

$\frac{A}{A_0} = e^{-\lambda t}$

$\frac{1.5}{15.0} = e^{-\lambda t}$

$0.1 = e^{-\lambda t}$

$\ln(0.1) = -\lambda t$

$-2.303 = -\lambda t$

$t = \frac{2.303}{\lambda}$

With $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5730} = 1.209 \times 10^{-4} \mathrm{ yr}^{-1}$:

$t = \frac{2.303}{1.209 \times 10^{-4}} = 19,040 \mathrm{ years}$

The ancient wood is approximately $19,000$ years old.

info

info many half-lives correspond to this fraction. $2^n = 10$ gives $n = \frac{\ln 10}{\ln 2} = 3.32$ half-lives. So $t = 3.32 \times 5730 = 19,024$ years. Both methods give the same result.

Worked Example 5: Nuclear Fission Energy

A nuclear power plant uses uranium-235 as fuel. Each fission of U-235 releases approximately $200$ MeV of energy. If the plant operates at a power output of $1000$ MW with an efficiency of $33\%$, calculate:

(a) The number of U-235 fissions per second (b) The mass of U-235 consumed per day

Solution

(a) Number of fissions per second:

The thermal power (energy per second from fission) is:

$P_{\mathrm{thermal}} = \frac{P_{\mathrm{electrical}}}{\mathrm{efficiency}} = \frac{1000 \times 10^6}{0.33} = 3.03 \times 10^9 \mathrm{ W}$

Energy released per fission:

$E_{\mathrm{fission}} = 200 \mathrm{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-11} \mathrm{ J}$

Number of fissions per second:

$\mathrm{Rate} = \frac{P_{\mathrm{thermal}}}{E_{\mathrm{fission}}} = \frac{3.03 \times 10^9}{3.2 \times 10^{-11}} = 9.47 \times 10^{19} \mathrm{ fissions/s}$

(b) Mass of U-235 consumed per day:

Number of fissions per day:

$N = 9.47 \times 10^{19} \times 86400 = 8.18 \times 10^{24} \mathrm{ fissions/day}$

Mass of U-235 per atom:

$m_{\mathrm{U-235}} = 235 \times 1.66 \times 10^{-27} = 3.90 \times 10^{-25} \mathrm{ kg}$

Mass consumed per day:

$m = N \times m_{\mathrm{U-235}} = 8.18 \times 10^{24} \times 3.90 \times 10^{-25} = 3.19 \mathrm{ kg/day}$

Worked Example 6: Activity and Number of Nuclei

A sample contains $5.0 \times 10^{20}$ atoms of cobalt-60 ($t_{1/2} = 5.27$ years). Calculate:

(a) The decay constant (b) The initial activity (c) The activity after 2 years

Solution

(a) Decay constant:

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5.27 \times 365.25 \times 24 \times 3600}$

$\lambda = \frac{0.693}{1.663 \times 10^8} = 4.17 \times 10^{-9} \mathrm{ s}^{-1}$

(b) Initial activity:

$A_0 = \lambda N_0 = 4.17 \times 10^{-9} \times 5.0 \times 10^{20} = 2.08 \times 10^{12} \mathrm{ Bq}$

(c) Activity after 2 years:

$t = 2 \times 365.25 \times 24 \times 3600 = 6.31 \times 10^7 \mathrm{ s}$

$A = A_0 e^{-\lambda t} = 2.08 \times 10^{12} \times e^{-(4.17 \times 10^{-9})(6.31 \times 10^7)}$

$A = 2.08 \times 10^{12} \times e^{-0.263}$

$A = 2.08 \times 10^{12} \times 0.769 = 1.60 \times 10^{12} \mathrm{ Bq}$

Worked Example 7: Penetration and Shielding

A gamma source emits radiation with a half-value thickness of $2.5$ cm in lead. How thick must the lead shield be to reduce the gamma intensity to $1/16$ of its original value?

Solution

If the intensity is reduced to $1/16$, this corresponds to $4$ half-value thicknesses (since $2^4 = 16$):

$\mathrm{Thickness} = 4 \times \mathrm{HVT} = 4 \times 2.5 = 10 \mathrm{ cm}$

Alternatively, using the exponential attenuation law:

$\frac{I}{I_0} = e^{-\mu x} = \frac{1}{16}$

$\mu = \frac{\ln 2}{\mathrm{HVT}} = \frac{0.693}{2.5} = 0.277 \mathrm{ cm}^{-1}$

$\frac{1}{16} = e^{-0.277x}$

$\ln\left(\frac{1}{16}\right) = -0.277x$

$-2.773 = -0.277x$

$x = 10.0 \mathrm{ cm}$

---

Common Mistakes

Mistake 1: Confusing Activity and Count Rate

Activity ($A$) is the number of decays per second from the source itself, measured in becquerels. Count rate ($R$) is the number of counts per second recorded by a detector.

$R = \eta \cdot A$

Where $\eta$ is the detection efficiency ($0 \lt \eta \lt 1$). In practice, $\eta$ depends on the solid angle subtended by the detector, absorption in air, the detector window, and the intrinsic efficiency of the detector.

Always use activity (not count rate) in decay law calculations. If a question gives count rate data, recognise that the count rate follows the same exponential decay pattern as activity, so you can still determine half-life from count rate measurements.

Mistake 2: Mixing Up Radiation Types

Confusion	Correct understanding
Alpha = helium-4 nucleus	$\prescript{4}{}{2}\mathrm{He}^{2+}$ (not just "helium")
Beta-minus = electron	Emitted from the nucleus (not an orbital electron)
Beta-plus = positron	Not the same as beta-minus; emitted by proton-rich nuclei
Gamma = photon	No charge, no mass; travels at $c$
Alpha has highest penetration	Wrong -- alpha has the LOWEST penetration (stopped by paper)
Beta has the lowest ionisation	Wrong -- gamma has the lowest ionisation power

Mistake 3: Binding Energy Sign Conventions

The binding energy is defined as a positive quantity. It represents the energy that must be supplied to separate the nucleus into its constituent nucleons.

$BE = (Zm_p + Nm_n - m_{\mathrm{nucleus}})c^2 \gt 0$

The mass defect $\Delta m$ is always positive for a bound nucleus:

$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}} \gt 0$

Do not write $BE = (m_{\mathrm{nucleus}} - Zm_p - Nm_n)c^2$ -- this would give a negative value, which is incorrect by definition.

Mistake 4: Forgetting Background Radiation

When measuring count rates with a GM tube, the measured count rate includes both the source and background radiation:

$R_{\mathrm{measured}} = R_{\mathrm{source}} + R_{\mathrm{background}}$

Always subtract the background count rate to obtain the true source count rate:

$R_{\mathrm{source}} = R_{\mathrm{measured}} - R_{\mathrm{background}}$

This is particularly important when the source count rate is comparable to the background rate.

Mistake 5: Incorrect Half-Life from Graph

When determining half-life from an exponential decay graph:

• The half-life is the time for the activity to drop to half its current value, not half the initial value after the first half-life. After the first half-life, the next half-life is measured from the current value, not from the initial value.
• On a log-linear plot, the slope is $-\lambda$, not $-t_{1/2}$. Use $t_{1/2} = \ln 2 / \lambda$.
• Ensure you are reading the correct values from the axes (check units).

Mistake 6: Not Balancing Decay Equations

Every decay equation must conserve:

• Mass number ($A$): sum of $A$ on left $=$ sum of $A$ on right
• Proton number ($Z$): sum of $Z$ on left $=$ sum of $Z$ on right
• Charge: sum of charges on left $=$ sum of charges on right

Common errors:

• Writing $A$ or $Z$ values incorrectly for daughter nuclei
• Forgetting that beta-minus decay increases $Z$ by $1$ (daughter is the next element)
• Forgetting that alpha decay decreases $Z$ by $2$ (daughter is two elements back)
• Confusing beta-minus (emits electron, $Z$ increases) with beta-plus (emits positron, $Z$ decreases)

Mistake 7: Using Electron Mass Instead of Atomic Mass

When calculating mass defects and binding energies, be consistent with the masses used:

• If using nuclear masses (mass of the bare nucleus): use $m_p$ for protons and $m_n$ for neutrons
• If using atomic masses (mass of the neutral atom): the atomic mass already includes the mass of the electrons, so use the atomic mass directly. The electron masses cancel out in the calculation

For most DSE problems, atomic masses are given, and the calculation simplifies because the electron masses cancel:

$\Delta m = Z \cdot m(\prescript{1}{}{1}\mathrm{H}) + N \cdot m_n - m(\prescript{A}{}{Z}\mathrm{X})$

Where $m(\prescript{1}{}{1}\mathrm{H})$ is the atomic mass of hydrogen (proton + electron).

warning

warning atomic masses are provided. Mixing the two conventions will lead to incorrect results. When in doubt, use atomic masses (the more common convention in exam questions) and note that the electron masses approximately cancel.

Mistake 8: Assuming All Radiation Is Equally Harmful

The biological effect of radiation depends on:

• The type of radiation (alpha is more damaging per unit dose but less penetrating)
• The dose received (higher dose = more damage)
• The duration of exposure
• Whether the source is external (alpha cannot penetrate skin) or internal (alpha is extremely dangerous if ingested)

An alpha source outside the body is relatively harmless (stopped by skin). An alpha source inside the body (e.g., inhaled or ingested) is extremely dangerous due to the high ionising power concentrated in a small volume of tissue.

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Problem Set

Problem 1: Balancing a Beta-Minus Decay Equation

Complete the following beta-minus decay equation and identify the daughter nucleus:

$\prescript{60}{}{27}\mathrm{Co} \to \ ? + e^- + \bar{\nu}_e$

Solution

In beta-minus decay, $Z$ increases by 1 while $A$ stays the same.

Daughter nucleus: $A = 60$, $Z = 27 + 1 = 28$

$\prescript{60}{}{27}\mathrm{Co} \to \prescript{60}{}{28}\mathrm{Ni} + e^- + \bar{\nu}_e$

The daughter is nickel-60 ($\prescript{60}{}{28}\mathrm{Ni}$).

If you get this wrong, revise: Beta-minus decay — $Z$ increases by 1, $A$ stays the same.

Problem 2: Activity from Mass

A sample of sodium-24 ($t_{1/2} = 15$ hours, molar mass $= 24 \mathrm{ g/mol}$) has a mass of $0.48 \mathrm{ g}$. Calculate its activity.

Solution

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{15 \times 3600} = 1.283 \times 10^{-5} \mathrm{ s}^{-1}$

$N = \frac{m}{M} N_A = \frac{0.48 \times 10^{-3}}{0.024} \times 6.02 \times 10^{23} = 0.02 \times 6.02 \times 10^{23} = 1.204 \times 10^{22}$

$A = \lambda N = (1.283 \times 10^{-5})(1.204 \times 10^{22}) = 1.545 \times 10^{17} \mathrm{ Bq}$

If you get this wrong, revise: Relationship between activity, decay constant, and number of nuclei ($A = \lambda N$) and $N = \frac{m}{M}N_A$.

Problem 3: Half-Life from Count Rate Data

A GM tube measures a count rate of $400$ counts/s from a radioactive source. After $20$ minutes, the count rate is $100$ counts/s. The background count rate is $20$ counts/s. Calculate the half-life of the source.

Solution

Subtract background:

Initial: $R_0 = 400 - 20 = 380$ counts/s

After 20 min: $R = 100 - 20 = 80$ counts/s

$\frac{R}{R_0} = \frac{80}{380} = 0.2105$

Using the decay law:

$0.2105 = e^{-\lambda \times 1200}$

$\lambda = \frac{-\ln(0.2105)}{1200} = \frac{1.558}{1200} = 1.298 \times 10^{-3} \mathrm{ s}^{-1}$

$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.298 \times 10^{-3}} = 533.9 \mathrm{ s} = 8.90 \mathrm{ minutes}$

If you get this wrong, revise: Background subtraction and half-life determination from count rate data.

Problem 4: Q-Value of a Nuclear Reaction

Calculate the Q-value of the fission reaction:

$\prescript{1}{}{0}\mathrm{n} + \prescript{235}{}{92}\mathrm{U} \to \prescript{144}{}{56}\mathrm{Ba} + \prescript{89}{}{36}\mathrm{Kr} + 3\prescript{1}{}{0}\mathrm{n}$

Given: $m(\mathrm{n}) = 1.008665$ u, $m(\mathrm{U\mathrm{-}235}) = 235.043930$ u, $m(\mathrm{Ba\mathrm{-}144}) = 143.922953$ u, $m(\mathrm{Kr\mathrm{-}89}) = 88.917630$ u.

Solution

$m_{\mathrm{reactants}} = 1.008665 + 235.043930 = 236.052595 \mathrm{ u}$

$m_{\mathrm{products}} = 143.922953 + 88.917630 + 3(1.008665) = 143.922953 + 88.917630 + 3.025995 = 235.866578 \mathrm{ u}$

$\Delta m = 236.052595 - 235.866578 = 0.186017 \mathrm{ u}$

$Q = 0.186017 \times 931.5 = 173.2 \mathrm{ MeV}$

If you get this wrong, revise: Q-value calculation — remember to account for all product particles including released neutrons.

Problem 5: Comparing Binding Energies

The binding energy per nucleon of deuterium ($\prescript{2}{}{1}\mathrm{H}$) is $1.11 \mathrm{ MeV/nucleon}$ and that of helium-4 ($\prescript{4}{}{2}\mathrm{He}$) is $7.07 \mathrm{ MeV/nucleon}$. Calculate the energy released when two deuterium nuclei fuse to form helium-4:

$2\prescript{2}{}{1}\mathrm{H} \to \prescript{4}{}{2}\mathrm{He}$

Solution

Total binding energy of reactants:

$BE_{\mathrm{reactants}} = 2 \times 2 \times 1.11 = 4.44 \mathrm{ MeV}$

Total binding energy of product:

$BE_{\mathrm{product}} = 4 \times 7.07 = 28.28 \mathrm{ MeV}$

Energy released:

$Q = BE_{\mathrm{product}} - BE_{\mathrm{reactants}} = 28.28 - 4.44 = 23.84 \mathrm{ MeV}$

If you get this wrong, revise: Binding energy per nucleon curve and how energy release relates to the increase in binding energy per nucleon.

Problem 6: Radiation Shielding — Multiple HVTs

A gamma source has a half-value thickness of $1.5 \mathrm{ cm}$ in concrete. What thickness of concrete is needed to reduce the intensity to $3\%$ of the original?

Solution

$\frac{I}{I_0} = 0.03$

$0.03 = \left(\frac{1}{2}\right)^n$

$n = \frac{\ln(1/0.03)}{\ln 2} = \frac{3.507}{0.693} = 5.06 \mathrm{ half-value thicknesses}$

$\mathrm{Thickness} = 5.06 \times 1.5 = 7.59 \mathrm{ cm} \approx 7.6 \mathrm{ cm}$

If you get this wrong, revise: Half-value thickness and exponential attenuation.

Problem 7: Fraction Remaining After Multiple Half-Lives

A radioactive isotope has a half-life of $8$ days. What fraction of the original sample remains after 30 days?

Solution

Number of half-lives: $n = \frac{30}{8} = 3.75$

$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{3.75} = \frac{1}{2^{3.75}} = \frac{1}{13.45} = 0.0743$

About $7.4\%$ of the original sample remains.

Alternatively, using the exponential decay law:

$\frac{N}{N_0} = e^{-\lambda t} = e^{-(\ln 2 / 8) \times 30} = e^{-2.601} = 0.0743$

If you get this wrong, revise: Fraction remaining after $n$ half-lives: $(1/2)^n$.

Problem 8: Dose Calculation

A patient receives a dose of $0.5 \mathrm{ mGy}$ from alpha radiation to a specific organ. Calculate the equivalent dose in mSv.

Solution

$H = D \times w_R = 0.5 \times 20 = 10 \mathrm{ mSv}$

If you get this wrong, revise: Absorbed dose vs equivalent dose and radiation weighting factors.

Problem 9: GM Tube Dead Time Correction

A GM tube with a dead time of $200 \ \mu\mathrm{s}$ records a count rate of $5000$ counts/s from a source. Estimate the true count rate.

Solution

The fraction of time the tube is dead:

$f = R_{\mathrm{measured}} \times \tau = 5000 \times 200 \times 10^{-6} = 1.0$

Since $f = 1.0$, the tube is dead $100\%$ of the time, which is physically impossible. This means the measured count rate of $5000$ counts/s is unreliable with this dead time.

For a more realistic scenario, if the measured rate were $1000$ counts/s:

$f = 1000 \times 200 \times 10^{-6} = 0.2$

$R_{\mathrm{true}} \approx \frac{R_{\mathrm{measured}}}{1 - f} = \frac{1000}{0.8} = 1250 \mathrm{ counts/s}$

If you get this wrong, revise: GM tube dead time and its effect on measured count rates at high activities.

Problem 10: Fusion vs Fission — Binding Energy Curve

Explain, with reference to the binding energy per nucleon curve, why energy is released in both nuclear fission and nuclear fusion.

Solution

The binding energy per nucleon curve has a peak around iron-56 ($\sim 8.8 \mathrm{ MeV/nucleon}$).

• Fusion: Light nuclei (low $A$) have low binding energy per nucleon. When they fuse to form heavier nuclei closer to the peak, the binding energy per nucleon increases. This means the products are more tightly bound than the reactants, so energy is released.

• Fission: Heavy nuclei (high $A$) have lower binding energy per nucleon than medium-mass nuclei near the peak. When a heavy nucleus splits into two lighter nuclei closer to the peak, the total binding energy increases. Energy is released because the products are more stable than the parent.

In both cases, energy is released because the products have a higher binding energy per nucleon (i.e., are more stable) than the reactants.

If you get this wrong, revise: The binding energy per nucleon curve and its interpretation.

Problem 11: Radiocarbon Dating — Percentage Remaining

An archaeological artefact has $35\%$ of the original C-14 remaining. How old is the artefact? ($t_{1/2}$ of C-14 $= 5730$ years)

Solution

$\frac{N}{N_0} = 0.35 = e^{-\lambda t}$

$\ln(0.35) = -\lambda t$

$t = \frac{-\ln(0.35)}{\lambda} = \frac{-(-1.050)}{(\ln 2)/5730} = \frac{1.050}{1.209 \times 10^{-4}} = 8683 \mathrm{ years}$

The artefact is approximately $8700$ years old.

If you get this wrong, revise: Radiocarbon dating formula and the decay law.

Problem 12: Moderator — Elastic Collision with Neutron

Explain why a moderator with a small mass number (like graphite or heavy water) is more effective at slowing neutrons than a heavy material like lead.

Solution

In an elastic head-on collision between a neutron (mass $m$) and a stationary nucleus (mass $M$), the fraction of kinetic energy transferred is:

$\frac{\Delta E_k}{E_k} = \frac{4mM}{(m + M)^2}$

This fraction is maximised when $M \approx m$ (i.e., when the moderator nucleus has a similar mass to the neutron).

For hydrogen ($M \approx m$): $\frac{\Delta E_k}{E_k} = \frac{4m^2}{(2m)^2} = 1$ (100% energy transfer)

For carbon-12 ($M = 12m$): $\frac{\Delta E_k}{E_k} = \frac{48m^2}{169m^2} = 0.284$ (28.4%)

For lead-207 ($M = 207m$): $\frac{\Delta E_k}{E_k} = \frac{828m^2}{43264m^2} = 0.019$ (1.9%)

A lighter moderator transfers much more energy per collision, so fewer collisions are needed to thermalise the neutrons.

If you get this wrong, revise: Nuclear reactor moderator and elastic collision energy transfer.

Problem 13: Alpha Particle Speed from Q-Value

In the alpha decay of Po-210 ($m = 209.9829$ u) to Pb-206 ($m = 205.9745$ u), the Q-value is $5.41 \mathrm{ MeV}$. Calculate the kinetic energy of the alpha particle ($m = 4.0026$ u).

Solution

By conservation of momentum, the alpha particle and daughter nucleus move in opposite directions with equal momentum:

$m_\alpha v_\alpha = m_{\mathrm{Pb}} v_{\mathrm{Pb}}$

The kinetic energies are in the inverse ratio of the masses:

$\frac{E_{k,\alpha}}{E_{k,\mathrm{Pb}}} = \frac{m_{\mathrm{Pb}}}{m_\alpha} = \frac{206}{4} = 51.5$

Since $E_{k,\alpha} + E_{k,\mathrm{Pb}} = Q = 5.41 \mathrm{ MeV}$:

$E_{k,\alpha} = Q \times \frac{m_{\mathrm{Pb}}}{m_{\mathrm{Pb}} + m_\alpha} = 5.41 \times \frac{206}{210} = 5.31 \mathrm{ MeV}$

The alpha particle carries approximately $98\%$ of the total kinetic energy.

If you get this wrong, revise: Conservation of momentum in nuclear decay and the distribution of kinetic energy between products.

Problem 14: Electron Capture

Write the equation for electron capture by beryllium-7 and explain why electron capture is favoured over beta-plus decay for this isotope.

Solution

$\prescript{7}{}{4}\mathrm{Be} + e^- \to \prescript{7}{}{3}\mathrm{Li} + \nu_e$

Electron capture is favoured because:

• Beta-plus decay requires $m(\mathrm{parent}) \gt m(\mathrm{daughter}) + 2m_e$, creating a positron and an additional electron
• Electron capture only requires $m(\mathrm{parent}) \gt m(\mathrm{daughter})$ (since the captured electron already exists)
• For light nuclei like Be-7, the mass difference is too small to create a positron, so electron capture is the only available decay mode

If you get this wrong, revise: Electron capture as an alternative to beta-plus decay and the condition $m(\mathrm{parent}) \gt m(\mathrm{daughter}) + 2m_e$.

Problem 15: Log-Linear Plot — Finding Decay Constant

A log-linear plot of $\ln A$ versus $t$ for a radioactive source gives a straight line with gradient $-0.05 \mathrm{ min}^{-1}$. Find the decay constant and half-life.

Solution

$\ln A = \ln A_0 - \lambda t$

The gradient is $-\lambda = -0.05 \mathrm{ min}^{-1}$

$\lambda = 0.05 \mathrm{ min}^{-1} = \frac{0.05}{60} = 8.33 \times 10^{-4} \mathrm{ s}^{-1}$

$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.05} = 13.86 \mathrm{ min}$

If you get this wrong, revise: Log-linear plots of radioactive decay — the gradient equals $-\lambda$, not $-t_{1/2}$.

For the A-Level treatment of this topic, see Radioactivity.

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tip

Diagnostic Test Ready to test your understanding of Nuclear Physics? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Nuclear Physics with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

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Derivations

Derivation: Radioactive Decay Law

The rate of decay of a radioactive sample is proportional to the number of undecayed nuclei:

$\frac{dN}{dt} = -\lambda N$

Separating variables and integrating:

$\int_{N_0}^{N} \frac{dN}{N} = \int_0^t -\lambda\, dt$

$\ln N - \ln N_0 = -\lambda t$

$N = N_0 e^{-\lambda t}$

Since activity $A = \lambda N$:

$A = A_0 e^{-\lambda t}$

Derivation: Half-Life Relation

At $t = t_{1/2}$, $N = N_0/2$:

$\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}}$

$\frac{1}{2} = e^{-\lambda t_{1/2}}$

$\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2}$

$-\ln 2 = -\lambda t_{1/2}$

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

Derivation: Mass-Energy Equivalence Applied to Nuclear Reactions

In a nuclear reaction, the total mass of the products differs from the total mass of the reactants. The mass defect $\Delta m$ corresponds to the energy released (or absorbed):

$\Delta E = \Delta m \cdot c^2$

For $1 \mathrm{ u}$ of mass defect:

$\Delta E = 1.661 \times 10^{-27} \times (3.0 \times 10^8)^2 = 1.495 \times 10^{-10} \mathrm{ J} = 931.5 \mathrm{ MeV}$

This is used to calculate binding energies and energy released in fission/fusion.

Derivation: Binding Energy per Nucleon

The binding energy of a nucleus with $Z$ protons and $N$ neutrons (mass number $A = Z + N$):

$B = [Zm_p + Nm_n - m_{\mathrm{nucleus}}]c^2$

where $m_{\mathrm{nucleus}}$ is the actual mass of the nucleus.

The binding energy per nucleon is $B/A$. Plotting $B/A$ versus $A$ shows:

• Light nuclei (low $A$): increasing $B/A$ (fusion releases energy).
• Iron-56 ($A = 56$): maximum $B/A$ (most stable nucleus).
• Heavy nuclei (high $A$): decreasing $B/A$ (fission releases energy).

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Experimental Methods

Investigating Radioactive Decay with a Geiger-Muller Tube

Apparatus: A Geiger-Muller (GM) tube connected to a counter/timer, a radioactive source (e.g., cobalt-60 or radon-220), and a ruler.

Procedure:

1. Place the GM tube at a fixed distance from the source.
2. Record the count rate at regular time intervals (e.g., every 30 seconds for radon-220, which has a short half-life of about 55 seconds).
3. Subtract the background count rate (measured with the source removed) from each reading.
4. Plot corrected count rate (or $\ln(\mathrm{count\ rate})$) versus time.
5. The half-life is determined from the time for the count rate to halve, or from the gradient of the ln(count rate) versus time graph.

Sources of error:

• Statistical fluctuations in radioactive decay (random nature).
• Background radiation changes during the experiment.
• Dead time of the GM tube (it cannot register counts during a brief recovery period after each detection).

Improvements: Take longer counting times to reduce statistical uncertainty. Repeat the experiment several times and average.

Determining the Half-Life of a Long-Lived Source

For a source with a half-life much longer than the practical measurement time, measure the activity at two widely separated times $t_1$ and $t_2$:

$A_1 = A_0 e^{-\lambda t_1}, \quad A_2 = A_0 e^{-\lambda t_2}$

$\frac{A_1}{A_2} = e^{-\lambda(t_1 - t_2)}$

$\lambda = \frac{\ln(A_1/A_2)}{t_2 - t_1}$

$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{(t_2 - t_1)\ln 2}{\ln(A_1/A_2)}$

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Data Analysis and Uncertainty

Statistical Uncertainty in Count Rate Measurements

Radioactive decay is a random process. The number of counts $N$ in time $t$ follows Poisson statistics. The standard deviation is $\sqrt{N}$.

For a count rate $R = N/t$:

$\Delta R = \frac{\sqrt{N}}{t} = \frac{\sqrt{Rt}}{t} = \sqrt{\frac{R}{t}}$

Example: A GM tube records $1200$ counts in $60 \mathrm{ s}$.

Count rate: $R = 1200/60 = 20 \mathrm{ counts/s}$

Uncertainty: $\Delta R = \sqrt{1200}/60 = 34.6/60 = 0.58 \mathrm{ counts/s}$

$R = (20.0 \pm 0.6) \mathrm{ counts/s}$

After subtracting background ($5.0 \pm 0.3 \mathrm{ counts/s}$):

$R_{\mathrm{net}} = (20.0 - 5.0) \pm \sqrt{0.6^2 + 0.3^2} = 15.0 \pm 0.67 \mathrm{ counts/s}$

$R_{\mathrm{net}} = (15.0 \pm 0.7) \mathrm{ counts/s}$

Analysing Decay Data with Logarithmic Plots

Plot $\ln A$ (y-axis) versus $t$ (x-axis). The equation $\ln A = \ln A_0 - \lambda t$ gives:

• Gradient $= -\lambda$
• Y-intercept $= \ln A_0$
• Half-life $= \ln 2 / |\mathrm{gradient}|$

The uncertainty in $\lambda$ is estimated from the worst-fit lines on the graph.

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Additional Worked Examples

Worked Example 11

A sample contains two radioactive isotopes: X (half-life $6.0 \mathrm{ hours}$, initial activity $800 \mathrm{ Bq}$) and Y (half-life $3.0 \mathrm{ hours}$, initial activity $400 \mathrm{ Bq}$). Calculate the total activity after $12 \mathrm{ hours}$.

Solution

For X: number of half-lives in 12 hours $= 12/6 = 2$

$A_X = 800 \times \left(\frac{1}{2}\right)^2 = 800 \times 0.25 = 200 \mathrm{ Bq}$

For Y: number of half-lives in 12 hours $= 12/3 = 4$

$A_Y = 400 \times \left(\frac{1}{2}\right)^4 = 400 \times 0.0625 = 25 \mathrm{ Bq}$

$A_{\mathrm{total}} = 200 + 25 = 225 \mathrm{ Bq}$

Worked Example 12

Calculate the energy released when a uranium-235 nucleus undergoes fission, given:

• Mass of U-235 $= 235.0439 \mathrm{ u}$
• Mass of Ba-141 $= 140.9139 \mathrm{ u}$
• Mass of Kr-92 $= 91.8973 \mathrm{ u}$
• Mass of 2 neutrons $= 2 \times 1.0087 = 2.0174 \mathrm{ u}$
• $1 \mathrm{ u} = 931.5 \mathrm{ MeV}/c^2$

Solution

$\Delta m = 235.0439 - (140.9139 + 91.8973 + 2.0174) = 235.0439 - 234.8286 = 0.2153 \mathrm{ u}$

$\Delta E = 0.2153 \times 931.5 = 200.6 \mathrm{ MeV}$

Worked Example 13

A nuclear power station produces $1500 \mathrm{ MW}$ of electrical power with an overall efficiency of $33\%$. Each fission of U-235 releases approximately $200 \mathrm{ MeV}$ of energy. Calculate the mass of U-235 consumed per day.

Solution

Total thermal power: $P_{\mathrm{thermal}} = 1500/0.33 = 4545 \mathrm{ MW} = 4.545 \times 10^9 \mathrm{ W}$

Energy per day: $E = 4.545 \times 10^9 \times 86400 = 3.93 \times 10^{14} \mathrm{ J}$

Energy per fission: $200 \mathrm{ MeV} = 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11} \mathrm{ J}$

Number of fissions per day: $N = \frac{3.93 \times 10^{14}}{3.2 \times 10^{-11}} = 1.23 \times 10^{25}$

Mass of U-235 per day: $m = N \times 235 \times 1.661 \times 10^{-27} = 1.23 \times 10^{25} \times 3.90 \times 10^{-25} = 4.80 \mathrm{ kg}$

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Exam-Style Questions

Question 1 (DSE Structured)

(a) Define the term "half-life".

(b) A radioactive isotope has a half-life of $8.0 \mathrm{ days}$. A sample initially contains $4.0 \times 10^{20}$ undecayed nuclei.

(i) Calculate the number of undecayed nuclei after $24 \mathrm{ days}$.

(ii) Calculate the decay constant.

(iii) Calculate the initial activity.

(iv) How long does it take for the activity to fall to $5.0 \times 10^8 \mathrm{ Bq}$?

Solution

(a) The half-life is the time taken for half of the radioactive nuclei in a sample to decay (or equivalently, for the activity to fall to half its initial value).

(b) (i) Number of half-lives in $24 \mathrm{ days}$: $n = 24/8 = 3$

$N = N_0 \times \left(\frac{1}{2}\right)^3 = 4.0 \times 10^{20} \times 0.125 = 5.0 \times 10^{19}$

(ii) $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.0 \times 24 \times 3600} = \frac{0.693}{691200} = 1.00 \times 10^{-6} \mathrm{ s}^{-1}$

(iii) $A_0 = \lambda N_0 = 1.00 \times 10^{-6} \times 4.0 \times 10^{20} = 4.0 \times 10^{14} \mathrm{ Bq}$

(iv) $A = 5.0 \times 10^8 = A_0 e^{-\lambda t} = 4.0 \times 10^{14} e^{-\lambda t}$

$e^{-\lambda t} = \frac{5.0 \times 10^8}{4.0 \times 10^{14}} = 1.25 \times 10^{-6}$

$-\lambda t = \ln(1.25 \times 10^{-6}) = -13.59$

$t = \frac{13.59}{1.00 \times 10^{-6}} = 1.36 \times 10^7 \mathrm{ s} = 157 \mathrm{ days}$

Question 2 (DSE Structured)

(a) Explain what is meant by "binding energy per nucleon".

(b) The following data are for several nuclei:

Nucleus	Mass (u)	Nucleon number $A$
H-2	2.01410	2
He-4	4.00260	4
C-12	12.00000	12
Fe-56	55.93494	56
U-235	235.04393	235

Mass of proton $= 1.00728 \mathrm{ u}$, mass of neutron $= 1.00867 \mathrm{ u}$.

(i) Calculate the binding energy of He-4 in MeV.

(ii) Calculate the binding energy per nucleon of He-4 and Fe-56.

(iii) Explain why energy is released when light nuclei undergo fusion and when heavy nuclei undergo fission.

Solution

(a) The binding energy per nucleon is the total binding energy of a nucleus divided by its mass number $A$. It represents the average energy needed to remove one nucleon from the nucleus. A higher binding energy per nucleon indicates greater nuclear stability.

(b) (i) He-4 has 2 protons and 2 neutrons.

$\Delta m = (2 \times 1.00728 + 2 \times 1.00867) - 4.00260 = 2.01456 + 2.01734 - 4.00260 = 0.03030 \mathrm{ u}$

$B = 0.03030 \times 931.5 = 28.2 \mathrm{ MeV}$

(ii) He-4: $B/A = 28.2/4 = 7.05 \mathrm{ MeV/nucleon}$

Fe-56 (26 protons, 30 neutrons):

$\Delta m = (26 \times 1.00728 + 30 \times 1.00867) - 55.93494 = 26.18928 + 30.26010 - 55.93494 = 0.51444 \mathrm{ u}$

$B = 0.51444 \times 931.5 = 479.2 \mathrm{ MeV}$

$$B/A = 479.2/56 = 8.56 \mathrm\\{ MeV/nucleon\\}$

(iii) The binding energy per nucleon curve peaks around Fe-56. Light nuclei (lower $B/A$) can increase their $B/A$ by fusing together (moving towards the peak), releasing energy equal to the difference in binding energies. Heavy nuclei (lower $B/A$ than the peak) can increase their $B/A$ by splitting apart (fission), also releasing energy. In both cases, the products have a higher binding energy per nucleon than the reactants, meaning they are more stable.

Question 3 (DSE Structured)

(a) Describe the operation of a Geiger-Muller tube.

(b) In a radiation experiment, a student measures the following count rates at different distances from a gamma source:

Distance $d$ (cm)	Count rate $R$ (counts/min)
2.0	3600
4.0	900
6.0	400
8.0	225
10.0	144

(i) Explain why the count rate decreases with distance.

(ii) Plot a suitable graph to verify the inverse square law.

(iii) The background count rate is $20 \mathrm{ counts/min}$. Calculate the corrected count rate at $d = 4.0 \mathrm{ cm}$ and the percentage correction.

Solution

(a) A Geiger-Muller tube consists of a metal cylinder (cathode) with a thin wire (anode) running along its axis, filled with an inert gas at low pressure. A high voltage is applied between the anode and cathode. When radiation enters the tube through a thin mica window, it ionises gas atoms. The ions are accelerated by the electric field, producing further ionisation (townsend avalanche). This creates a pulse of current that is registered as one count. The quenching gas (e.g., halogen) absorbs UV photons to prevent secondary discharges.

(b) (i) Gamma radiation obeys the inverse square law: $R \propto 1/d^2$. As the distance increases, the radiation is spread over a larger area, so the count rate decreases.

(ii) Plot $R$ (y-axis) versus $1/d^2$ (x-axis):

$1/d^2$ (cm$^{-2}$)	$R$ (counts/min)
0.250	3600
0.0625	900
0.0278	400
0.0156	225
0.0100	144

The graph is approximately a straight line through the origin, confirming the inverse square law.

(iii) Corrected count rate at $d = 4.0 \mathrm{ cm}$: $R_{\mathrm{corrected}} = 900 - 20 = 880 \mathrm{ counts/min}$

Percentage correction: $\frac{20}{900} \times 100\% = 2.2\%$

The background correction is small at close range but becomes more significant at larger distances where the count rate is lower.