IB DP Physics · Theme E · Nuclear and quantum physics

Radioactive decay.

E.3 · Binding energy, the strong nuclear force, alpha, beta and gamma decay, half-life
Standard level and higher level

SL + HL · 7 hours Additional HL · 5 hours Nuclear Decay sim ↗

The journey

Four parts.

Part 01 · SL + HL

What holds a nucleus together?

The strong force, mass defect, binding energy

Part 02 · SL + HL

How does a nucleus change?

Alpha, beta and gamma decay

Part 03 · SL + HL

How fast does it happen?

Activity, half-life, background

Part 04 · HL only

Stability and the decay law

Energy levels, the neutrino, \(N = N_{0} e^{-\lambda t}\)

Slides with a terracotta edge and an HL only flag are additional higher level. Every other slide is for SL and HL.

E.3 RADIOACTIVE DECAY · SL + HL

Part 01 · SL + HL

What holds a nucleus together?

Isotopes, the strong nuclear force, mass defect and binding energy

E.3 RADIOACTIVE DECAY · SL + HL

Same element, different nuclei

Isotopes.

THREE ISOTOPES OF CARBON

Same number of protons: the same element, the same chemistry
Different numbers of neutrons: a different mass
A nuclide is one kind of nucleus: \({}^{14}_{6}\mathrm{C}\) has \(A = 14\) and \(Z = 6\)
Some isotopes are stable, others are radioactive

E.3 RADIOACTIVE DECAY · SL + HL

Why the chemistry is the same

Chemical behavior is decided by the electrons, and a neutral atom has as many electrons as it has protons. Extra neutrons change the mass of the nucleus but not its charge, so all three kinds of carbon form the same compounds. Carbon dioxide made with carbon-14 behaves in the atmosphere, and in a plant, exactly like ordinary carbon dioxide.

Why the nuclear behavior differs

Stability depends on the balance of protons and neutrons in the nucleus. Carbon-12 and carbon-13 are stable. Carbon-14 has too many neutrons for its six protons and decays, with a half-life of 5730 years. This is what makes radiocarbon dating possible, later in the deck.

Nucleon number \(A\) = protons + neutrons. Proton number \(Z\) = protons. Neutrons = \(A - Z\). E.1 introduced this notation.

Holding the protons in

The strong nuclear force.

Protons repel one another, yet nuclei exist
An attractive force acts between nucleons: proton or neutron alike
Short range: only between near neighbors
Much stronger than the electric force at that range

The curves are not required: know that the strong force is attractive and short range.

TWO PROTONSSKETCH, NOT TO SCALE

E.3 RADIOACTIVE DECAY · SL + HL

How big the problem is

Two protons 2 fm apart repel with \(\frac{8.99 \times 10^{9} \times (1.60 \times 10^{-19})^{2}}{(2 \times 10^{-15})^{2}} = 58 \,\mathrm{N}\). That is the weight of a 6 kg mass, acting on a particle of mass \(1.7 \times 10^{-27} \,\mathrm{kg}\). Gravity between them is weaker by a factor of about \(10^{36}\). Something far stronger must hold them, and it must act on neutrons too, since neutrons stay in the nucleus.

Why short range matters

A nucleon feels the strong force only from the nucleons touching it, but a proton feels the electric repulsion of every other proton in the nucleus. In a large nucleus the repulsion adds up while the attraction does not. That is why heavy nuclei need extra neutrons, and why the very heaviest are unstable.

At very small separations, well under 1 fm, the strong force becomes repulsive. This stops the nucleus collapsing and is why all nuclei have about the same density (E.1, HL).

Mass and energy

Mass defect.

\(E = mc^{2}\)

A nucleus has less mass than its separate nucleons
The difference is the mass defect
The missing mass was released as energy when the nucleus formed

MAKING HELIUM-4

E.3 RADIOACTIVE DECAY · SL + HL

What \(E = mc^{2}\) says

Mass and energy are two measures of the same thing, linked by the factor \(c^{2}\). When a system gives out energy \(E\), its mass falls by \(\frac{E}{c^{2}}\). This happens in a chemical reaction too, but there the change is about one part in \(10^{10}\) and cannot be measured. In nuclear changes it is nearly one percent, and it can.

Three units of mass

Nuclear masses are given in kilograms, in unified atomic mass units (u), or in \(\mathrm{MeV} \,c^{-2}\). The last one makes the arithmetic disappear: a mass of \(1 \,\mathrm{u}\) is equivalent to \(931.5 \,\mathrm{MeV}\) of energy, so \(1 \,\mathrm{u} = 931.5 \,\mathrm{MeV} \,c^{-2}\). A proton has a mass of \(938 \,\mathrm{MeV} \,c^{-2}\).

\(1 \,\mathrm{u} = 1.661 \times 10^{-27} \,\mathrm{kg}\) is one twelfth of the mass of a carbon-12 atom. All of these values are in the data booklet.

How tightly a nucleus is held

Nuclear binding energy.

The energy needed to separate a nucleus completely into its nucleons
Equal to the energy released when the nucleus forms
Binding energy = mass defect × \(c^{2}\)

Divide by \(A\) for the binding energy per nucleon: a measure of stability.

Helium-4 nucleus: \(4.001506 \,\mathrm{u}\) Proton \(1.007276 \,\mathrm{u}\) · neutron \(1.008665 \,\mathrm{u}\) \(\Delta m = m_\mathrm{He} - (2 m_\mathrm{p} + 2 m_\mathrm{n})\) \(\Delta m = 4.001506 - 4.031882 = -0.030376 \,\mathrm{u}\) \(E_\mathrm{b} = -\Delta m\,c^{2} = -(-0.030376) \times 931.5 = \class{res}{28.3 \,\mathrm{MeV}}\) Per nucleon: \(\dfrac{28.3}{4} = \class{res}{7.07 \,\mathrm{MeV}}\)

E.3 RADIOACTIVE DECAY · SL + HL

Binding energy is not energy the nucleus has

For formation from separate nucleons, \(\Delta m = m_\mathrm{final} - m_\mathrm{initial} < 0\). The mass defect is the positive magnitude \(-\Delta m\), and the positive binding energy is \(E_\mathrm{b} = -\Delta m c^{2}\). It is energy the nucleus has lost: the amount you would have to put back in to pull every nucleon free. A nucleus with a large binding energy sits deep in an energy well and is hard to break up. Think of it as a debt, not as savings.

Why divide by the number of nucleons

A big nucleus has a large total binding energy simply because it has many nucleons. To compare how firmly each nucleon is held in different nuclei, divide by \(A\). The higher the binding energy per nucleon, the more stable the nucleus.

Rounding the masses early. The mass defect is a small difference between large numbers: keep every digit you are given until the subtraction is done.

The most important graph in nuclear physics

The binding energy curve.

Rises steeply for light nuclei
Peaks near iron-56: the most stable nuclei, about 8.8 MeV
Falls slowly for heavy nuclei

Any change that moves up the curve releases energy.

BINDING ENERGY PER NUCLEONMEASURED VALUES

E.3 RADIOACTIVE DECAY · SL + HL

Fusion and fission on one graph

Joining two light nuclei gives a product further to the right and higher up: each nucleon is more tightly bound than before, so energy is released. That is fusion, the source of the Sun’s energy. Splitting a very heavy nucleus gives two products further to the left, also higher up: fission. Both move towards iron, from opposite ends.

How much energy

The energy released is the total binding energy of the products minus that of what you started with. When uranium-235 (7.6 MeV per nucleon) splits into fragments at about 8.5 MeV per nucleon, each of the 235 nucleons gains about 0.9 MeV: roughly 200 MeV for one fission, millions of times more than any chemical reaction between two atoms.

The spike at helium-4

Helium-4 is far more tightly bound than its neighbors. This is why alpha particles, and not single protons or neutrons, are what heavy nuclei throw out.

Saying iron has the most binding energy. It has the most binding energy per nucleon. Uranium has a far larger total.

Worked example

The energy released when radium-226 decays.

\({}^{226}_{88}\mathrm{Ra}\) → \({}^{222}_{86}\mathrm{Rn}\) + \(\class{alpha}{{}^{4}_{2}\mathrm{He}}\)

Masses: \(226.025410 \,\mathrm{u}\) → \(222.017578 \,\mathrm{u}\) + \(4.002603 \,\mathrm{u}\) \(\Delta m = m_\mathrm{after} - m_\mathrm{before}\) \(\Delta m = 226.020181 - 226.025410 = -0.005229 \,\mathrm{u}\) \(E_\mathrm{released} = -\Delta m\,c^{2} = -(-0.005229) \times 931.5 = \class{res}{4.87 \,\mathrm{MeV}}\)

The products have less mass: energy is released
It appears as kinetic energy, mostly of the alpha particle
Released energy: \(7.8 \times 10^{-13} \,\mathrm{J}\)

E.3 RADIOACTIVE DECAY · SL + HL

Why the alpha particle takes most of the energy

The radium nucleus is at rest, so the total momentum afterwards must be zero: the alpha particle and the radon nucleus fly apart with equal and opposite momenta (A.2). Kinetic energy is \(\frac{p^{2}}{2m}\), so for the same momentum the lighter particle has more of it. The alpha particle, with 4 of the 226 mass units, carries \(\frac{222}{226}\) of the energy: 4.78 MeV.

A decay can only happen if mass is lost

If the products would have more mass than the original nucleus, the decay cannot happen by itself, because it would need energy from nowhere. This test tells you which decays are possible. The rest-energy change is \(\Delta E_\mathrm{rest} = \Delta m c^{2} < 0\); the products gain kinetic energy, so \(\Delta E_\mathrm{rest} + \Delta E_\mathrm{k} = 0\).

These are masses of whole atoms, electrons included. The 88 electrons on the left balance the 86 + 2 on the right, so they cancel.

Quick check

Binding energy.

01How do isotopes of an element differ?

In their number of neutrons

02Give two properties of the strong nuclear force.

Attractive between nucleons, short range

03What is the mass defect of a nucleus?

Mass of its separate nucleons minus its own mass

04A mass defect of 0.100 u. Binding energy?

\(0.100 \times 931.5 = 93.2 \,\mathrm{MeV}\)

05Which nuclei have the greatest binding energy per nucleon?

Those near iron-56

06Why does fusing light nuclei release energy?

The product is higher on the curve

PRESS → TO REVEAL EACH ANSWER

Part 02 · SL + HL

How does a nucleus change?

Alpha, beta and gamma decay, the neutrino, and what each radiation does to matter

E.3 RADIOACTIVE DECAY · SL + HL

The story · Paris, 1896

Background story · not assessed

Becquerel’s fogged plate.

Photographic portrait of Henri Becquerel, a bearded man in a dark suit, seated — **Henri Becquerel**, photographed by Paul Nadar

A photographic plate with two dark smudges and handwritten notes above them. In the lower smudge the pale outline of a cross can be seen — His **photographic plate**, darkened by uranium salts

X-rays had just been discovered. Becquerel thought sunlight made uranium salts give them out
He left the salts on a wrapped photographic plate in a dark drawer
The plate darkened anyway, with no sunlight
The radiation came from the uranium itself. Nobel Prize with the Curies, 1903

Spontaneous: nothing was done to the uranium to make it happen.

Portrait: Paul Nadar, public domain. Plate: Henri Becquerel, 1896, public domain.

E.3 RADIOACTIVE DECAY · SL + HL

The nature of radioactive decay

Random and spontaneous.

An unstable nucleus changes into a more stable one and emits radiation

Random

Impossible to predict

Nobody can say which nucleus will decay next, or when a particular one will. Each has the same chance in each second

Spontaneous

Impossible to influence

Temperature, pressure and chemical state make no difference. The decay happens by itself

Random for one nucleus, yet very predictable for a large number.

E.3 RADIOACTIVE DECAY · SL + HL

How something random can be predictable

Nobody can say whether a single coin will land heads. Toss a million coins and you can be sure that very close to half a million will. A sample of radioactive material holds an enormous number of nuclei, so the fraction that decays in a given time is extremely steady, even though each decay is a matter of chance. That steady fraction is what half-life measures.

Why chemistry cannot touch it

Heating, squeezing or reacting a substance changes how its electrons and atoms behave. The energies involved are a few electronvolts. Changes inside a nucleus involve millions of electronvolts, so the nucleus does not notice what the electrons are doing.

Evidence for randomness: the count rate from a steady source is never perfectly steady. It fluctuates from one second to the next.

Emitting a helium nucleus

Alpha decay.

An alpha particle is a helium-4 nucleus: 2 protons and 2 neutrons
\(A\) falls by 4, \(Z\) falls by 2
Typical of heavy nuclei

\({}^{241}_{95}\mathrm{Am}\)→\({}^{237}_{93}\mathrm{Np}\)+\(\class{alpha}{{}^{4}_{2}\mathrm{He}}\)

AMERICIUM-241PRESS → TO DECAY

E.3 RADIOACTIVE DECAY · SL + HL

Why heavy nuclei do this

A nucleus with more than 82 protons is too large for the short-range strong force to hold against the long-range repulsion of all its protons. Throwing out two protons and two neutrons together reduces the charge and moves the nucleus up the binding energy curve. The helium-4 group is so tightly bound itself that emitting it as one piece releases far more energy than emitting four separate nucleons would.

One energy

All the alpha particles from a given decay come out with the same kinetic energy, typically 4 to 9 MeV. That sharp energy is evidence that nuclei have discrete energy levels, which the HL part returns to.

In any decay equation the top numbers balance and the bottom numbers balance: \(241 = 237 + 4\) and \(95 = 93 + 2\).

A neutron becomes a proton

Beta-minus decay.

A neutron changes into a proton, an electron and an antineutrino
\(A\) unchanged, \(Z\) rises by 1
Nuclei with too many neutrons

\({}^{14}_{6}\mathrm{C}\)→\({}^{14}_{7}\mathrm{N}\)+\(\class{electron}{{}^{0}_{-1}\mathrm{e}}\)+\(\bar{\nu}\)

CARBON-14PRESS → FOR EACH STAGE

E.3 RADIOACTIVE DECAY · SL + HL

The electron was not there before

There are no electrons inside a nucleus. The beta particle is created at the instant of the decay, when one of the neutrons turns into a proton, and it leaves at once. It is an ordinary electron, but it comes from the nucleus, not from the electron shells of the atom.

Why the bottom number is −1

The electron is written \({}^{0}_{-1}\mathrm{e}\): nucleon number 0, because it is not a nucleon, and “proton number” −1, standing for its charge. Then both rows balance: \(14 = 14 + 0\) and \(6 = 7 - 1\). Charge is conserved: one positive proton and one negative electron appear together.

Saying the nucleus loses a neutron and therefore has a smaller nucleon number. It has one neutron fewer and one proton more: the total is unchanged.

A proton becomes a neutron

Beta-plus decay.

A proton changes into a neutron, a positron and a neutrino
\(A\) unchanged, \(Z\) falls by 1
Nuclei with too few neutrons

\({}^{18}_{9}\mathrm{F}\)→\({}^{18}_{8}\mathrm{O}\)+\(\class{electron}{{}^{0}_{+1}\mathrm{e}}\)+\(\nu\)

FLUORINE-18PRESS → FOR EACH STAGE

E.3 RADIOACTIVE DECAY · SL + HL

The positron

A positron is the antiparticle of the electron: the same mass, the opposite charge. It does not last long. Within a fraction of a millimeter in a solid it meets an electron and the two annihilate, their mass turning into two gamma photons that travel in opposite directions.

Where this is used

Fluorine-18 is the isotope used in PET scans. It is attached to a sugar that collects in active tissue. Detectors on opposite sides of the patient pick up the pairs of gamma photons, and a computer works back to where each pair came from.

Only inside a nucleus

A free proton never decays like this, because a neutron has more mass than a proton. Inside a nucleus the energy can come from the rest of the nucleus becoming more tightly bound.

Minus goes with antineutrino, plus with neutrino. In each decay one particle and one antiparticle are created.

The third particle

Neutrinos and antineutrinos.

Every beta decay also emits a neutrino \(\nu\) or an antineutrino \(\bar{\nu}\)

Charge

None

Not affected by electric or magnetic fields

Mass

Almost none

Less than about a millionth of the mass of an electron

Interaction

Hardly any

Passes through the whole Earth almost unhindered

It carries away energy and momentum. Without it, beta decay would break the conservation laws.

E.3 RADIOACTIVE DECAY · SL + HL

HL only How anyone knows it is there

When a nucleus at rest emits a single particle, the two must fly apart back to back and share the energy in a fixed way. In beta decay the electron and the recoiling nucleus are often seen not moving back to back, and the electron’s energy varies from one decay to the next. Something unseen must be taking the missing momentum and energy.

How many there are

The Sun’s fusion reactions make neutrinos in vast numbers. About \(6 \times 10^{10}\) of them pass through every square centimeter of you each second, day and night, since the Earth does not stop them. Almost none interact.

The force responsible for beta decay, the weak nuclear force, is not part of this course. You need the neutrino’s existence and its place in the equations.

Losing energy, not particles

Gamma decay.

A nucleus left excited gives out its extra energy as a photon
\(A\) and \(Z\) do not change
Usually follows an alpha or beta decay

\({}^{99\mathrm{m}}_{43}\mathrm{Tc}\)→\({}^{99}_{43}\mathrm{Tc}\)+\(\class{gamma}{\gamma}\)

TECHNETIUM-99mPRESS → TO DECAY

E.3 RADIOACTIVE DECAY · SL + HL

The same idea as atomic spectra

In E.1 an electron dropping to a lower level in an atom emits a photon whose energy equals the difference between the levels. A nucleus also has energy levels. After an alpha or beta decay the new nucleus is often left in an excited level, and it drops to its ground state by emitting a photon. The difference is the size of the energy: a few eV for an atom, hundreds of keV or several MeV for a nucleus.

Photons from a nucleus have much shorter wavelengths

A 1.0 MeV gamma photon has \(\lambda = \frac{hc}{E} = \frac{1.24 \times 10^{-6}}{1.0 \times 10^{6}} = 1.2 \times 10^{-12} \,\mathrm{m}\), about half a million times shorter than visible light.

The “m” in technetium-99m means metastable: an excited state that lasts unusually long, hours in this case and not a tiny fraction of a second.

Balancing the numbers

Decay equations.

Decay	Emitted	\(A\)	\(Z\)	Example
Alpha	\(\class{alpha}{{}^{4}_{2}\mathrm{He}}\)	−4	−2	\({}^{238}_{92}\mathrm{U}\) → \({}^{234}_{90}\mathrm{Th}\) + \(\class{alpha}{{}^{4}_{2}\mathrm{He}}\)
Beta-minus	\(\class{electron}{{}^{0}_{-1}\mathrm{e}}\) + \(\bar{\nu}\)	0	+1	\({}^{90}_{38}\mathrm{Sr}\) → \({}^{90}_{39}\mathrm{Y}\) + \(\class{electron}{{}^{0}_{-1}\mathrm{e}}\) + \(\bar{\nu}\)
Beta-plus	\(\class{electron}{{}^{0}_{+1}\mathrm{e}}\) + \(\nu\)	0	−1	\({}^{22}_{11}\mathrm{Na}\) → \({}^{22}_{10}\mathrm{Ne}\) + \(\class{electron}{{}^{0}_{+1}\mathrm{e}}\) + \(\nu\)
Gamma	\(\class{gamma}{\gamma}\)	0	0	\({}^{60}_{28}\mathrm{Ni}\)* → \({}^{60}_{28}\mathrm{Ni}\) + \(\class{gamma}{\gamma}\)

E.3 RADIOACTIVE DECAY · SL + HL

Two conservation laws do all the work

The top numbers must add up to the same total on both sides: nucleons are neither created nor destroyed. The bottom numbers must add up too: that is conservation of charge. Neutrinos and photons have zero for both, so they never affect the balance.

Working out an unknown

Polonium-210 (\(Z = 84\)) decays by alpha emission. The new nucleus has \(A = 210 - 4 = 206\) and \(Z = 84 - 2 = 82\). Look up \(Z = 82\): lead. So the product is lead-206. Always find the element from the proton number.

The star on nickel-60 marks an excited state. It was left that way by the beta decay of cobalt-60, which is why cobalt-60 is used as a gamma source.

Leaving out the neutrino or antineutrino. It is needed for full marks in a beta decay equation.

What the radiation does to matter

Penetrating and ionizing ability.

WHAT STOPS EACH RADIATIONPRESS → FOR EACH RADIATION

The more strongly a radiation ionizes, the sooner it loses its energy and stops.

E.3 RADIOACTIVE DECAY · SL + HL

Ionizing and penetrating are opposites

Radiation loses energy each time it knocks an electron out of an atom. An alpha particle is slow, heavy and has charge \(+2e\): it ionizes thousands of atoms in every millimeter of air, so its energy is used up within a few centimeters. A beta particle is fast and light with charge \(e\), ionizes far less often, and travels meters. A gamma photon has no charge and ionizes only when it happens to hit an electron, so it can pass through centimeters of lead.

Gamma radiation is never completely stopped

Each centimeter of lead removes the same fraction of the photons. The intensity keeps falling, but it never reaches exactly zero, which is why the diagram says “reduced”.

The danger follows from this. Outside the body, gamma is the most hazardous because it gets in. Inside the body, alpha is the worst, because all its energy is delivered to a tiny volume of tissue.

Side by side

Comparing alpha, beta and gamma.

	Alpha	Beta	Gamma
What it is	Helium-4 nucleus	Electron or positron	Photon
Charge	\(+2e\)	\(-e\) or \(+e\)	0
Ionizing ability	Very strong	Moderate	Weak
Range in air	A few cm	About a meter	Very long
Stopped by	Paper, or skin	A few mm of aluminum	Reduced by cm of lead

E.3 RADIOACTIVE DECAY · SL + HL

Telling them apart in the laboratory

Put absorbers between a source and a detector. If a sheet of paper cuts the count rate sharply, alpha is present. If paper makes little difference but a few millimeters of aluminum cuts it, beta. If the count rate is still well above background behind aluminum, gamma.

Using a field

In a magnetic field alpha and beta-minus particles curve in opposite directions, because their charges have opposite signs, and gamma goes straight on. The beta particles curve much more sharply than the alpha particles: they have a far smaller mass, so much less momentum for the same sort of energy (D.3).

Choosing a source for a job means matching this table to the task: what must the radiation pass through, and what must stop it?

The story · Paris, from 1898

Background story · not assessed

Marie Curie’s notebooks.

Portrait of Marie Curie, an older woman with her hair tied back — **Marie Curie**, pictured about 1920

An open notebook with squared yellowed pages covered in handwriting and small sketches of apparatus — One of her **laboratory notebooks**, now in London. Tests found radium on it

She and her husband Pierre discovered polonium and radium
Nobody knew the danger. She kept tubes of radioactive material in her pocket
Radioactive atoms got onto her things. Her notebooks are still contaminated

Radium-226 has a half-life of about 1600 years: they will stay radioactive for well over 1000 more.

Portrait: Henri Manuel, Wellcome Collection, CC BY 4.0 (cropped). Notebook: Wellcome Collection, CC BY 4.0.

E.3 RADIOACTIVE DECAY · SL + HL

Quick check

Radioactive decay.

01What does “spontaneous” mean for radioactive decay?

It cannot be influenced from outside

02\({}^{226}_{88}\mathrm{Ra}\) emits an alpha particle. \(A\) and \(Z\) of the product?

222 and 86

03What is emitted with the electron in beta-minus decay?

An antineutrino

04\({}^{14}_{6}\mathrm{C}\) decays by beta-minus. \(A\) and \(Z\) of the product?

14 and 7

05Which radiation is the most ionizing?

Alpha

06What happens to \(A\) and \(Z\) in gamma decay?

Neither changes

PRESS → TO REVEAL EACH ANSWER

Part 03 · SL + HL

How fast does it happen?

Activity, count rate, half-life and background radiation

E.3 RADIOACTIVE DECAY · SL + HL

Measuring decay

Activity and count rate.

Activity

Decays per second in the source

Measured in becquerels: 1 Bq = 1 decay per second. A property of the sample

Count rate

Counts per second at a detector

Only the radiation that reaches the detector and is recorded. Depends on the set-up

With a fixed arrangement, the count rate is proportional to the activity
Both are proportional to the number of undecayed nuclei in the sample

E.3 RADIOACTIVE DECAY · SL + HL

Why the count rate is smaller than the activity

A source sends radiation out in all directions and the detector covers only a small part of the sphere round it. Some radiation is absorbed by the air or by the source itself on the way, and some passes through the detector without being recorded. A source of activity 40 000 Bq might give a count rate of a few hundred counts per second.

Why that does not matter for half-life

If nothing in the arrangement is moved, the detector always catches the same fraction. When the activity halves, the count rate halves. So the half-life can be found from the count rate without ever knowing the activity itself.

Activity is proportional to the number of undecayed nuclei because each nucleus has the same chance of decaying in each second: twice as many nuclei, twice as many decays.

The time to halve

Half-life.

The time for half the undecayed nuclei in a sample to decay
Also the time for the activity, or the count rate, to halve
Always the same for a given nuclide, whenever you start

After \(n\) half-lives: \(\left(\dfrac{1}{2}\right)^{n}\) remains.

HALF-LIFE 8 DAYSPRESS → FOR EACH HALF-LIFE

E.3 RADIOACTIVE DECAY · SL + HL

Why it halves and does not fall steadily

Each nucleus has the same chance of decaying in the next second, however long it has already survived. With half as many nuclei left, half as many decay each second. So the sample loses the same fraction in equal times, not the same number: a half in the first 8 days, a half of what is left in the next 8 days.

It never quite reaches zero

After 10 half-lives about one thousandth remains, \(\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\). After 20, about one millionth. Half-lives range from a tiny fraction of a second to billions of years: 4.5 billion years for uranium-238, about the age of the Earth.

Thinking everything has decayed after two half-lives. After two, a quarter remains.

Chance, in large numbers

Random decay, predictable half-life.

64 NUCLEIPRESS → FOR EACH HALF-LIFE

undecayeddecayed

About half decay in each half-life, but not exactly half, and nobody can say which ones.

E.3 RADIOACTIVE DECAY · SL + HL

What the grids show

With only 64 nuclei, chance matters: 33 survive the first half-life where 32 were expected, and 15 survive the second where 16 or 17 were expected. Run it again and the numbers would be slightly different, with different nuclei surviving.

Why real samples behave so regularly

The scatter is roughly the square root of the number expected. For 32 it is about 6, nearly 20%. For a real sample with \(10^{20}\) nuclei it is about \(10^{10}\), one part in ten billion. The randomness is still there, but it is far too small to see.

A good classroom model: throw 100 dice, remove every six, and repeat. The number left falls with a “half-life” of just under four throws.

Worked example

Counting half-lives.

Activity 800 Bq · half-life 6.0 hours Activity after 24 hours? Number of half-lives \(n = \dfrac{t}{T_{\frac{1}{2}}} = \dfrac{24}{6.0} = 4\) \(A = A_{0} \left(\dfrac{1}{2}\right)^{n} = 800 \times \left(\dfrac{1}{2}\right)^{4}\) \(A = \class{res}{50 \,\mathrm{Bq}}\)

A sample falls to \(\frac{1}{8}\) of its activity in 15 days \(\dfrac{A}{A_{0}} = \left(\dfrac{1}{2}\right)^{n} = \dfrac{1}{8}\) so \(n = 3\) \(T_{\frac{1}{2}} = \dfrac{15}{3} = \class{res}{5.0 \text{ days}}\)

E.3 RADIOACTIVE DECAY · SL + HL

Radiation that is always there

Background radiation.

A detector counts even with no source present
From rocks and soil, radon gas, cosmic rays, food, medical uses
Subtract it from every reading

Corrected count rate = measured − background.

Background: 110 counts in 5 minutes Background rate \(= \dfrac{110}{5} = 22\) counts per minute With the source: 126 counts per minute Corrected \(= 126 - 22 = \class{res}{104}\) counts per minute

E.3 RADIOACTIVE DECAY · SL + HL

Why measure the background for a long time

The background count rate is low and random, so a short measurement is unreliable: 20 counts in one minute could easily have been 15 or 25. Counting for several minutes and dividing gives a much steadier average.

When it matters most

For a strong source the background is a small correction. But as the source decays, its count rate falls towards the background, and in the end most of what the detector records has nothing to do with the source. Without the correction, the count rate would seem to level off above zero and the half-life would come out too long.

Measure the background with the source well away, in the same place and with the same detector, before or after the experiment.

From a graph

Finding a half-life from a graph.

Subtract the background first
Plot corrected count rate against time and draw a smooth curve
Read off the time for the count rate to halve
Repeat from another point and average

HALF-LIFE 2.0 HOURS

E.3 RADIOACTIVE DECAY · SL + HL

Reading this graph

The corrected count rate starts at 480 counts per minute. It reaches 240 after 2.0 hours and 120 after 4.0 hours: two readings of the half-life, both 2.0 hours. Starting from 400 and reading the time to reach 200 would give a third. Agreement between them is the check that the decay really is exponential.

What goes wrong without the correction

The measured curve starts at 510 and would need to fall to 255. That takes about 2.2 hours, and the next halving takes longer still. A “half-life” that keeps growing is the sign that the background has not been removed.

Draw the curve through the scatter of the points, not from point to point. Each reading is uncertain because decay is random.

Choosing the right isotope

Putting radioactivity to work.

Medical tracer

Gamma · half-life of hours

Must leave the body to reach a camera, and be gone within a day or so. Technetium-99m: 6 hours

Leaks in underground pipes

Gamma · half-life of hours or days

Must pass through soil to a detector above, and not linger in the water supply

Thickness of materials

Beta · half-life of years

Partly absorbed by paper or foil, so the count rate shows the thickness. Must stay steady

Radioactive dating

Half-life similar to the age

Carbon-14, 5730 years, for once-living material. Uranium-238 for rocks

E.3 RADIOACTIVE DECAY · SL + HL

Two questions decide every choice

First, penetration: what does the radiation have to get through, and what should it not get through? Alpha would never leave the body or the pipe. Gamma would pass through a sheet of paper as if it were not there, so it could not measure its thickness. Second, half-life: long enough to do the job, short enough not to remain a hazard afterwards.

How carbon dating works

Cosmic rays keep making carbon-14 in the atmosphere, and living things take it in, so while they are alive they contain a steady, known proportion. At death the intake stops and the carbon-14 decays. The fraction left tells you how long ago that was. After about ten half-lives too little is left to measure.

A sample has a quarter of the carbon-14 of living wood. A quarter is two half-lives: \(2 \times 5730 = 11\,460\) years old.

Quick check

Half-life.

01What is the unit of activity?

The becquerel, Bq: one decay per second

02What fraction of a sample remains after 3 half-lives?

\(\frac{1}{8}\)

03Activity falls from 1200 Bq to 150 Bq in 12 hours. Half-life?

3 half-lives, so 4.0 hours

04A detector reads 95 counts per minute; background is 15. Corrected count rate?

80 counts per minute

05Which radiation would suit a gauge for the thickness of paper?

Beta

06Why does a medical tracer need a short half-life?

So the patient’s dose is small

PRESS → TO REVEAL EACH ANSWER

Part 04 · Additional higher level

Stability and the decay law.

Evidence for the strong force · neutrons and stability
Nuclear energy levels · the neutrino · \(N = N_{0} e^{-\lambda t}\)

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

How do we know?

Evidence for the strong nuclear force.

Nuclei exist

Protons stay together

An attraction stronger than their electric repulsion must act at nuclear distances

Scattering

Rutherford’s law fails

Alpha particles that come very close to a nucleus are not scattered as the electric force alone predicts (E.1)

Binding energy

Roughly constant per nucleon

About 8 MeV for every nucleon in a medium or heavy nucleus: each binds only to its neighbors

Together: a force that is strong, attractive, short range and the same for protons and neutrons.

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

Why constant binding energy means short range

Suppose every nucleon attracted every other nucleon, as every mass attracts every other mass. The number of pairs among \(A\) nucleons grows as \(A^{2}\), so the binding energy per nucleon would grow in proportion to \(A\). It does not: above \(A = 60\) it stays close to 8 MeV. So each nucleon must be bound only to the few nucleons next to it. Adding more nucleons to the far side of the nucleus makes no difference to it. The force is said to saturate.

Why the curve still droops

The electric repulsion is long range: every proton does repel every other proton. That energy grows faster than the number of nucleons and slowly lowers the binding energy per nucleon for heavy nuclei, from 8.8 MeV at iron to 7.6 MeV at uranium.

A liquid behaves the same way: each molecule is held only by its neighbors, so the energy needed to boil off a gram is the same whatever the size of the drop.

The ratio of neutrons to protons

Neutrons and stability.

Light stable nuclei: \(N \approx Z\)
Heavy stable nuclei need more neutrons: about 1.5 for each proton
Off the band: beta decay moves a nucleus back towards it
Beyond lead, \(Z = 82\): no stable nuclides

THE BAND OF STABILITY

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

What the neutrons do

A neutron adds strong attraction without adding any electric repulsion. It also spaces the protons further apart. In a small nucleus equal numbers are enough. As \(Z\) grows, the total repulsion grows quickly, and more and more neutrons are needed to hold the protons in: lead-208 has 82 protons and 126 neutrons.

How decay follows the chart

A nucleus above the band has too many neutrons. Beta-minus decay turns one neutron into a proton: one step down and one step right, towards the band. A nucleus below the band has too few. Beta-plus decay turns a proton into a neutron: one step up and one left. Alpha decay removes two of each, moving a heavy nucleus diagonally down the chart.

Carbon-14 has \(N = 8\), \(Z = 6\): above the band, so beta-minus. Fluorine-18 has \(N = 9\), \(Z = 9\): below the band for its size, so beta-plus.

Evidence from alpha and gamma spectra

Discrete nuclear energy levels.

Alpha particles from one nuclide have a few sharp energies
Gamma photons have sharp energies too
The differences between the alpha energies match the gamma energies

A nucleus, like an atom, has discrete energy levels.

ONE PARENT, TWO ROUTES

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

The argument

The parent nucleus starts with a fixed energy. If the daughter could be left with any energy at all, the alpha particles would come out with a continuous spread. They do not: their energies are a short list of exact values. So the daughter can be left only in certain states: its ground state, or one of a few excited states, each with a definite energy.

The check

A daughter left in an excited state then drops to the ground state and emits a gamma photon. Its energy should equal the difference between the two alpha routes, apart from the small recoil of the nucleus. Measurements agree: most alpha particles from americium-241 have about 60 keV less than the most energetic ones, and the decay is followed by a 59.5 keV gamma photon.

Compare E.1: line spectra are the evidence for discrete energy levels in atoms. Nuclear levels are about a million times further apart.

A spectrum that should not be continuous

The beta spectrum and the neutrino.

Beta particles have a continuous range of energies, up to a maximum
Yet the nucleus changes between two fixed levels
The rest of the energy goes to a third particle: the neutrino

\(E_\mathrm{max}\) is the full energy of the decay.

BETA PARTICLES FROM ONE NUCLIDE

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

Why two particles would give one energy

If the decay produced only the daughter nucleus and an electron, conservation of momentum and energy together would fix how the energy is shared, as in alpha decay. Every electron would come out with the same energy, close to the maximum. The measured spread means that energy and momentum are being shared among three bodies, and three bodies can share them in endless ways.

Reading the graph

An electron with nearly \(E_\mathrm{max}\) left almost nothing for the neutrino. An electron with very little energy came from a decay in which the neutrino took nearly all of it. In every decay the electron and the neutrino together carry \(E_\mathrm{max}\).

Contrast the three spectra: alpha and gamma are discrete, evidence for nuclear energy levels. Beta is continuous, evidence for the neutrino.

The story · 1930 to 1956

Background story · not assessed

The particle nobody could find.

Photographic portrait of Wolfgang Pauli, a man with dark hair in a suit and tie — **Wolfgang Pauli**, 1945

Two men in shirts and ties working at a large bank of electronic instruments covered in dials and cables — **Clyde Cowan** and **Frederick Reines** with their detector’s electronics

The beta spectrum seemed to break conservation of energy
1930: Pauli proposed a light, neutral particle in a letter that began “Dear radioactive ladies and gentlemen”
1956: Cowan and Reines detected antineutrinos from a nuclear reactor at Savannah River
Nobel Prize for Reines in 1995

Trusting a conservation law predicted a new particle.

Pauli: Nobel Foundation, 1945, public domain. Cowan and Reines: Savannah River Site, public domain.

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

The chance of decay

The decay constant and the decay law.

\(N = N_{0} e^{-\lambda t}\)

\(\lambda\): the decay constant, in s⁻¹
The probability that a nucleus decays in unit time
\(N\): undecayed nuclei left after time \(t\)

Where it comes from

Decays per second \(\propto\) nuclei left

A quantity that falls at a rate proportional to itself falls exponentially

A limit on “probability”

Only when \(\lambda t\) is small

\(\lambda \times t\) is the chance of decay in time \(t\) only if it is much less than 1

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

What \(\lambda\) means

If \(\lambda = 0.010 \,\mathrm{s}^{-1}\), each nucleus has a 1% chance of decaying in the next second, and in a large sample 1% of the nuclei do decay in that second. A large \(\lambda\) means a short-lived nuclide.

Why only for small \(\lambda t\)

Take iodine-131, with \(\lambda = 0.0864\) per day. Over one day, \(\lambda t = 0.086\) and the true fraction that decays is \(1 - e^{-0.0864} = 0.083\): close. Over 20 days, \(\lambda t = 1.73\), which cannot be a probability at all, since it is more than 1. The true fraction decayed is \(1 - e^{-1.73} = 0.82\). The simple product works only while the number of nuclei has hardly changed during the interval.

\(\lambda\) and \(t\) must use the same unit of time: per second with seconds, per day with days. Their product has no unit.

Three linked equations

Activity and half-life.

\(A = \lambda N = \lambda N_{0} e^{-\lambda t}\)

Activity is the rate of decay
It falls with the same exponential as \(N\): \(A = A_{0} e^{-\lambda t}\)

At \(t = T_{\frac{1}{2}}\), half the nuclei are left \(\dfrac{1}{2} N_{0} = N_{0} e^{-\lambda T_{\frac{1}{2}}}\) \(\ln 2 = \lambda T_{\frac{1}{2}}\) \(T_{\frac{1}{2}} = \dfrac{\ln 2}{\lambda}\)

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

Why \(A = \lambda N\)

Each of the \(N\) nuclei has a probability \(\lambda\) of decaying in each second, so the number of decays per second is \(\lambda N\). This one line connects something you can measure with a detector, the activity, to something you cannot count directly, the number of nuclei.

The half-life step

Cancel \(N_{0}\) to get \(\frac{1}{2} = e^{-\lambda T}\). Take natural logarithms: \(\ln \frac{1}{2} = -\lambda T\). Since \(\ln \frac{1}{2} = -\ln 2\), the minus signs cancel. \(\ln 2 = 0.693\), so a quick form is \(T_{\frac{1}{2}} = \frac{0.693}{\lambda}\).

A long half-life means a small decay constant and, for the same number of nuclei, a low activity. Uranium-238 is only weakly radioactive because its half-life is so long.

Worked example

Iodine-131 after 20 days.

Half-life 8.02 days · initial activity 400 kBq \(\lambda = \dfrac{\ln 2}{T_{\frac{1}{2}}} = \dfrac{0.693}{8.02} = 0.0864 \text{ day}^{-1}\) \(A = A_{0} e^{-\lambda t}\) \(A = 400 \times e^{-0.0864 \times 20} = 400 \times 0.178\) \(A = \class{res}{71 \,\mathrm{kBq}}\)

20 days is 2.5 half-lives: not a whole number
Check: between \(\frac{1}{4}\) (100 kBq) and \(\frac{1}{8}\) (50 kBq)
Time in days, so \(\lambda\) in day⁻¹

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

Worked example

The activity of 1.0 μg of cobalt-60.

Half-life 5.27 years \(= 1.66 \times 10^{8} \,\mathrm{s}\) \(\lambda = \dfrac{\ln 2}{T_{\frac{1}{2}}} = \dfrac{0.693}{1.66 \times 10^{8}} = 4.17 \times 10^{-9} \,\mathrm{s}^{-1}\) Mass of one atom ≈ \(60 \,\mathrm{u}\) \(N = \dfrac{m}{m_\mathrm{atom}} = \dfrac{1.0 \times 10^{-9}}{60 \times 1.661 \times 10^{-27}} = 1.00 \times 10^{16}\) \(A = \lambda N = 4.17 \times 10^{-9} \times 1.00 \times 10^{16}\) \(A = \class{res}{4.2 \times 10^{7} \,\mathrm{Bq}}\)

For becquerels, \(\lambda\) must be in s⁻¹
\(N\) from the mass and the mass of one atom
A millionth of a gram: forty million decays every second

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

Measuring a very long half-life

This calculation can be run backwards. Nobody can watch uranium-238 for 4.5 billion years. But you can weigh a sample, work out \(N\), measure its activity \(A\) with a detector, and find \(\lambda = \frac{A}{N}\). The half-life follows from \(\frac{\ln 2}{\lambda}\). Long half-lives are measured this way, without waiting for any noticeable change.

Using moles

The number of atoms can also be found as \(\frac{\text{mass}}{\text{molar mass}} \times N_\mathrm{A}\): \(\frac{1.0 \times 10^{-6} \,\mathrm{g}}{60 \,\mathrm{g} \,\mathrm{mol}^{-1}} \times 6.02 \times 10^{23} = 1.0 \times 10^{16}\). Both routes give the same number.

One year is \(3.16 \times 10^{7} \,\mathrm{s}\). It is worth remembering for this topic and for D.1.

Finding \(\lambda\) from data

A straight-line graph for decay.

\(A = A_{0} e^{-\lambda t}\) \(\ln A = \ln A_{0} - \lambda t\)

Plot \(\ln A\) against \(t\): a straight line
Gradient \(= -\lambda\)
Then \(T_{\frac{1}{2}} = \dfrac{\ln 2}{\lambda}\)

LOGARITHM OF ACTIVITY AGAINST TIME

E.3 RADIOACTIVE DECAY · ADDITIONAL HL

Why bother

Judging a half-life from a curve depends on how well the curve has been drawn through scattered points. A straight line is much easier to judge by eye, and its gradient uses every reading at once. If the points do not lie on a straight line, the decay is not a single exponential: perhaps the background was not subtracted, or there are two nuclides in the sample.

A case with numbers

If \(\ln A\) falls from 6.0 to 3.0 while \(t\) goes from 0 to 25 minutes, the gradient is \(\frac{3.0 - 6.0}{25} = -0.12 \,\mathrm{min}^{-1}\). So \(\lambda = 0.12 \,\mathrm{min}^{-1}\) and \(T_{\frac{1}{2}} = \frac{0.693}{0.12} = 5.8\) minutes.

Radioactive dating at higher level

A piece of wood has 35% of the carbon-14 activity of living wood. \(\lambda = \frac{0.693}{5730} = 1.21 \times 10^{-4} \,\mathrm{year}^{-1}\), and \(t = \frac{\ln \left(\frac{1}{0.35}\right)}{\lambda} = \frac{1.05}{1.21 \times 10^{-4}} = 8700\) years.

The corrected count rate can be used in place of the activity. It only shifts the line up or down, and the gradient is the same.

Quick check · additional higher level

Stability and the decay law.

01Why do heavy stable nuclei need extra neutrons?

More attraction, no more repulsion

02Which spectra are evidence for nuclear energy levels?

Alpha and gamma: both discrete

03What does the continuous beta spectrum show?

A third particle: the neutrino

04A nuclide has \(\lambda = 0.050 \,\mathrm{s}^{-1}\). Half-life?

\(\frac{0.693}{0.050} = 14 \,\mathrm{s}\)

05\(N = 2.0 \times 10^{12}\), \(\lambda = 1.5 \times 10^{-6} \,\mathrm{s}^{-1}\). Activity?

\(3.0 \times 10^{6} \,\mathrm{Bq}\)

06What is the gradient of a graph of \(\ln A\) against \(t\)?

\(-\lambda\)

PRESS → TO REVEAL EACH ANSWER

Summary

Radioactive decay on one slide.

Binding energy

Mass defect × \(c^{2}\)

\(1 \,\mathrm{u} = 931.5 \,\mathrm{MeV} \,c^{-2}\)
Per nucleon: greatest near iron-56

Strong nuclear force

Short range, attractive

Between all nucleons
Balances the repulsion of the protons

Decay

Random and spontaneous

Alpha, beta-minus with \(\bar{\nu}\), beta-plus with \(\nu\), gamma
\(A\) and \(Z\) balance

Half-life

The time to halve

\(\left(\frac{1}{2}\right)^{n}\) remains after \(n\) half-lives
Subtract the background

Evidence · HL

Spectra

Alpha and gamma discrete: nuclear levels
Beta continuous: the neutrino

Decay law · HL

\(N = N_{0} e^{-\lambda t}\)

\(A = \lambda N\) and \(T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}\)

E.3 RADIOACTIVE DECAY · SL + HL

Summary

Key words.

IsotopesSame protons, different neutrons

Mass defectMass lost when a nucleus forms

Binding energyEnergy to separate every nucleon

Strong nuclear forceShort-range attraction of nucleons

SpontaneousNot affected by outside conditions

AntineutrinoEmitted in beta-minus decay

ActivityDecays per second, in Bq

Half-lifeTime for half the nuclei to decay

Background radiationAlways present. Subtract it

Decay constantProbability of decay per unit time

Band of stabilityStable nuclides on an N–Z chart

Continuous spectrumEvery energy up to a maximum

E.3 RADIOACTIVE DECAY · SL + HL