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Radioactive decay, half-life and uses

Radioactive decay, half-life and uses

6.3.1 Beta-minus and beta-plus decay

Beta decay in unstable nuclei

Definition

Beta particle

A beta particle is an electron emitted from the nucleus of an unstable atom.

Definition

Positron

A positron is a particle with the same relative mass as an electron and a relative charge of +1.

  1. A nucleus is held together by the balance between the strong nuclear force pulling all the nucleons together and the electrostatic repulsion pushing the protons apart.
  2. When a nucleus holds too many neutrons or too many protons for that balance to hold, it is unstable and it will eventually decay.
  3. Beta decay is the route an unstable nucleus takes when its problem is the ratio of neutrons to protons rather than its total size, so no nucleons are thrown out of the nucleus at all.
  4. Instead, one nucleon changes identity inside the nucleus, and a small light particle is created and thrown out to carry away the charge.
  5. There are two versions of this, and which one happens depends on which nucleon the nucleus has too many of.
    1. A nucleus with too many neutrons turns a neutron into a proton and emits an electron, which is called beta-minus decay and written β−\beta^-β− decay.
    2. A nucleus with too many protons turns a proton into a neutron and emits a positron, which is called beta-plus decay and written β+\beta^+β+ decay.
  6. The two emitted particles are antiparticles of each other, so they have the same very small relative mass but opposite relative charge.

Beta-minus decay

Definition

Beta-minus decay

Beta-minus decay is the decay of an unstable nucleus in which a neutron becomes a proton and an electron, and the electron is emitted from the nucleus as a beta-minus particle.

  1. Inside a neutron-rich nucleus, a single neutron changes into a proton and an electron, and the electron is immediately expelled from the nucleus.
  2. In words, the change to the nucleon is neutron→proton+electron\text{neutron} \rightarrow \text{proton} + \text{electron}neutron→proton+electron.
  3. Using particle symbols, the same change is n→p+e−\text{n} \rightarrow \text{p} + \text{e}^-n→p+e−.
  4. Written with mass numbers and charges, so that both quantities can be seen to balance, the change is 01n→11p+−10e_{0}^{1}\text{n} \rightarrow _{1}^{1}\text{p} + _{-1}^{0}\text{e}01​n→11​p+−10​e.
  5. The emitted electron is the beta-minus particle, and in a nuclear equation it is written as −10e_{-1}^{0}\text{e}−10​e or −10β_{-1}^{0}\beta−10​β.
  6. Both symbols carry the same information: a mass number of 000, because an electron contains no protons or neutrons, and a charge of −1-1−1, because it carries one electron charge.
  7. The electron that leaves is created in the nucleus at the moment of decay, and it is not one of the electrons that was orbiting the atom beforehand.
  8. Beta-minus emitters are common among the heavy nuclei produced in reactors and among the daughter products of long decay chains, because those nuclei are typically left neutron-rich.

Diagram of the beta-minus decay of thorium-234. A neutron in the nucleus changes into a proton, an electron is emitted from the nucleus, and the atomic number rises from 90 to 91 to give protactinium-234 while the mass number stays at 234.

Example

Carbon-14 decaying by beta-minus emission

  • Carbon-14 is written 614C_{6}^{14}\text{C}614​C, so it has 666 protons and 14−6=814 - 6 = 814−6=8 neutrons.
  • Eight neutrons to six protons is a neutron-rich mixture for such a light nucleus, so this nucleus decays by β−\beta^-β− emission.
  • One neutron becomes a proton, so the proton count rises from 666 to 777 while the total nucleon count stays at 141414.
  • An element with 777 protons is nitrogen, so the daughter nucleus is 714N_{7}^{14}\text{N}714​N.
  • The decay is written 614C→714N+−10e_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e}614​C→714​N+−10​e, with the emitted electron shown as a product.
  • Nothing was ejected from the nucleus except that electron, so the same fourteen nucleons are present throughout.

Beta-plus decay

Definition

Beta-plus decay

Beta-plus decay is the decay of an unstable nucleus in which a proton becomes a neutron and a positron, and the positron is emitted from the nucleus.

  1. Inside a proton-rich nucleus, a single proton changes into a neutron and a positron, and the positron is immediately expelled from the nucleus.
  2. In words, the change to the nucleon is proton→neutron+positron\text{proton} \rightarrow \text{neutron} + \text{positron}proton→neutron+positron.
  3. Using particle symbols, the same change is p→n+e+\text{p} \rightarrow \text{n} + \text{e}^+p→n+e+.
  4. Written with mass numbers and charges, the change is 11p→01n++10e_{1}^{1}\text{p} \rightarrow _{0}^{1}\text{n} + _{+1}^{0}\text{e}11​p→01​n++10​e.
  5. The emitted positron is written +10e_{+1}^{0}\text{e}+10​e or +10β_{+1}^{0}\beta+10​β, with a mass number of 000 and a charge of +1+1+1.
  6. A positron is a form of antimatter, so once it has left the nucleus it travels only a very short distance through matter before it meets an ordinary electron and the pair is annihilated.
  7. Because a positron is destroyed so quickly, beta-plus emitters are much harder to keep and store than beta-minus emitters, and they are usually made artificially.
Example

Fluorine-18 decaying by beta-plus emission

  • Fluorine-18 is written 918F_{9}^{18}\text{F}918​F, so it has 999 protons and 18−9=918 - 9 = 918−9=9 neutrons.
  • Nine protons to nine neutrons leaves this nucleus proton-rich compared with the stable isotope fluorine-19, so it decays by β+\beta^+β+ emission.
  • One proton becomes a neutron, so the proton count falls from 999 to 888 while the total nucleon count stays at 181818.
  • An element with 888 protons is oxygen, so the daughter nucleus is 818O_{8}^{18}\text{O}818​O.
  • The decay is written 918F→818O++10e_{9}^{18}\text{F} \rightarrow _{8}^{18}\text{O} + _{+1}^{0}\text{e}918​F→818​O++10​e, with the emitted positron shown as a product.
  • The positron leaves almost at once and is annihilated, so what remains in the sample is oxygen-18.

Comparing the two beta decays

  1. Both decays start with an unstable nucleus, change one nucleon into the other kind of nucleon, and emit one light particle from the nucleus.
  2. The differences all follow from which nucleon changes.
    1. In β−\beta^-β− decay a neutron is the nucleon that changes, so the neutron count falls by one and the proton count rises by one.
    2. In β+\beta^+β+ decay a proton is the nucleon that changes, so the proton count falls by one and the neutron count rises by one.
  3. The emitted particles differ only in the sign of their charge, which is what the minus and plus in the two names record.
  4. Neither decay ejects a nucleon from the nucleus, which is what separates beta decay from alpha decay and from neutron emission.
  5. Neither decay can be triggered or prevented, because which nucleus decays and when is settled by chance alone.
Exam technique

Describing a beta decay for full marks

  • Name the nucleon that changes and the nucleon it becomes, because a description that only names the emitted particle has left out the change inside the nucleus.
  • Say that the light particle is emitted from the nucleus, since the mark is for the emission and not just for the particle existing.
  • Write β−\beta^-β− decay as a neutron changing into a proton with an electron emitted, and β+\beta^+β+ decay as a proton changing into a neutron with a positron emitted.
  • Match the sign to the particle every time: minus goes with the electron, −10e_{-1}^{0}\text{e}−10​e, and plus goes with the positron, +10e_{+1}^{0}\text{e}+10​e.
  • Use the word nucleus rather than atom or particle when you say where the emission comes from, because the looser word is not credited.
Common Mistake
  • Do not describe the beta-minus particle as an electron that has escaped from an orbit, because it is created inside the nucleus when a neutron changes.
  • Do not write that a positron is an emitted proton, because the proton stays in the nucleus as a neutron and the positron is a separate new particle.
  • Do not give a beta particle a mass number of 111, since its mass number is 000 and only protons and neutrons are counted in a mass number.
  • Do not let the mass number change in a beta decay, because the total number of nucleons is unchanged by either process.
  • Do not confuse a positron with a positive ion, because an ion is an atom that has lost or gained electrons and is far more massive.
Self review
  • What happens to a neutron during β−\beta^-β− decay, and which particle is emitted?
  • What happens to a proton during β+\beta^+β+ decay, and which particle is emitted?
  • Give the mass number and the charge of a positron.
  • Why does neither beta decay eject a nucleon from the nucleus?
  • Write the particle equation for the nucleon change in β+\beta^+β+ decay.
  • Explain why the electron emitted in β−\beta^-β− decay is not one of the atom's orbiting electrons.

6.3.2 Effects of decay on the nucleus

The two numbers that describe a nucleus

Definition

Atomic number

The number of protons in the nucleus of an atom.

Definition

Mass number

The total number of protons and neutrons in the nucleus of an atom.

Definition

Daughter nucleus

A daughter nucleus is the nucleus left behind after a parent nucleus has decayed.

  1. A nucleus is written in the form ZAX_{Z}^{A}\text{X}ZA​X, where X\text{X}X is the chemical symbol, AAA is the mass number and ZZZ is the atomic number.
  2. The atomic number is the count of protons, and it alone decides which element the nucleus belongs to, so any decay that changes ZZZ changes the element.
  3. The mass number is the count of protons plus neutrons, so the neutron count of a nucleus is always A−ZA - ZA−Z.
  4. Every radioactive decay removes something from the nucleus or converts one nucleon into another, so every decay leaves AAA and ZZZ changed in a way fixed by what was emitted.
  5. The nucleus that is left behind after a decay is the daughter nucleus, and the nucleus that decayed is called the parent nucleus.
  6. Because both numbers are conserved across a decay, working out the daughter is a matter of subtracting the mass number and the charge of the emitted particle from the parent.

Alpha decay

Definition

Alpha particle

An alpha particle is a particle made of two protons and two neutrons emitted from an unstable nucleus, and is equivalent to a helium nucleus.

  1. An alpha particle is a tightly bound group of two protons and two neutrons, identical to a helium nucleus, and it is written 24α_{2}^{4}\alpha24​α or 24He_{2}^{4}\text{He}24​He.
  2. Alpha decay happens in very heavy nuclei, where the electrostatic repulsion between a large number of protons makes the whole nucleus too large to hold together.
  3. Ejecting an alpha particle removes four nucleons at once, which is the fastest way for such a nucleus to shed both size and positive charge.
  4. The alpha particle carries away a mass number of 444, so the mass number of the nucleus falls by 444.
  5. It also carries away a charge of +2+2+2, so the atomic number falls by 222.
  6. Since the atomic number has changed, the daughter nucleus is a different element, two places back in the periodic table.
  7. In general form, alpha decay is ZAX→Z−2A−4Y+24α_{Z}^{A}\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\alphaZA​X→Z−2A−4​Y+24​α.

Diagram of the alpha decay of uranium-235 into thorium-231 and an alpha particle. The mass number falls from 235 to 231 and the atomic number falls from 92 to 90, because the alpha particle carries away two protons and two neutrons.

Example

Finding the daughter of an alpha emitter

  • Radium-226 is written 88226Ra_{88}^{226}\text{Ra}88226​Ra, so it has 888888 protons and 226−88=138226 - 88 = 138226−88=138 neutrons.
  • The mass number of the daughter is 226−4=222226 - 4 = 222226−4=222.
  • The atomic number of the daughter is 88−2=8688 - 2 = 8688−2=86.
  • The element with 868686 protons is radon, so the daughter nucleus is 86222Rn_{86}^{222}\text{Rn}86222​Rn.
  • The complete equation is 88226Ra→86222Rn+24α_{88}^{226}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\alpha88226​Ra→86222​Rn+24​α.
  • The neutron count has fallen from 138138138 to 222−86=136222 - 86 = 136222−86=136, which confirms that two protons and two neutrons left together.

Beta-minus and beta-plus decay

  1. In beta-minus decay a neutron becomes a proton and an electron is emitted, and the electron is written −10e_{-1}^{0}\text{e}−10​e.
  2. Because the electron has a mass number of 000, the total nucleon count is unchanged, so the mass number stays the same.
  3. Because a neutron has been replaced by a proton, the atomic number rises by 111, and the daughter is the next element up.
  4. In general form, beta-minus decay is ZAX→Z+1AY+−10e_{Z}^{A}\text{X} \rightarrow _{Z+1}^{A}\text{Y} + _{-1}^{0}\text{e}ZA​X→Z+1A​Y+−10​e, as in 614C→714N+−10e_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e}614​C→714​N+−10​e.
  5. In beta-plus decay a proton becomes a neutron and a positron is emitted, and the positron is written +10e_{+1}^{0}\text{e}+10​e.
  6. The mass number again stays the same, for the same reason, but the atomic number falls by 111, so the daughter is the previous element.
  7. In general form, beta-plus decay is ZAX→Z−1AY++10e_{Z}^{A}\text{X} \rightarrow _{Z-1}^{A}\text{Y} + _{+1}^{0}\text{e}ZA​X→Z−1A​Y++10​e, as in 1122Na→1022Ne++10e_{11}^{22}\text{Na} \rightarrow _{10}^{22}\text{Ne} + _{+1}^{0}\text{e}1122​Na→1022​Ne++10​e.
  8. Both beta decays therefore change the element without changing the mass of the nucleus in any measurable way, because the neutron count and proton count trade places.

Gamma emission and nuclear rearrangement

Definition

Gamma radiation

Gamma radiation is electromagnetic radiation emitted from the nucleus of an unstable atom when that nucleus loses energy.

Definition

Nuclear rearrangement

Nuclear rearrangement is the movement of the protons and neutrons of a nucleus into a lower energy arrangement after a decay, with the surplus energy released as gamma radiation.

  1. Gamma radiation is not a particle at all; it is a very high frequency electromagnetic wave carrying energy away from the nucleus.
  2. A gamma ray has no mass number and no charge, so it is written 00γ_{0}^{0}\gamma00​γ.
  3. The direct consequence is that after gamma emission the mass number stays the same, the atomic number stays the same, and the element does not change.
  4. What does change is the energy of the nucleus, which is left lower than it was before.
  5. Gamma emission normally follows another decay rather than happening on its own, because an alpha or beta decay usually leaves the daughter nucleus with its nucleons in a high energy arrangement.
  6. That high energy state is described as an excited nucleus, and it settles by nuclear rearrangement: the protons and neutrons shift into a lower energy arrangement and the surplus energy leaves as a gamma ray.
  7. Because gamma emission needs no nucleon to leave and no nucleon to change, it is the only one of these processes that leaves the identity of the nucleus completely untouched.
  8. In general form, gamma emission is ZAX∗→ZAX+00γ_{Z}^{A}\text{X}^{*} \rightarrow _{Z}^{A}\text{X} + _{0}^{0}\gammaZA​X∗→ZA​X+00​γ, where the asterisk marks the excited nucleus.

Diagram of thorium-234 decaying by beta-minus emission to an excited protactinium-234 nucleus, which then emits a gamma photon to reach a lower energy state without any change to its mass number or atomic number.

Example

A beta decay followed by gamma emission

  • Thorium-234, written 90234Th_{90}^{234}\text{Th}90234​Th, is neutron-rich and decays by beta-minus emission.
  • The mass number stays at 234234234 and the atomic number rises from 909090 to 919191, giving protactinium-234.
  • The equation for the first stage is 90234Th→91234Pa+−10e_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}\text{e}90234​Th→91234​Pa+−10​e.
  • The protactinium nucleus is left excited, so its nucleons rearrange and it emits a gamma ray.
  • The equation for the second stage is 91234Pa∗→91234Pa+00γ_{91}^{234}\text{Pa}^{*} \rightarrow _{91}^{234}\text{Pa} + _{0}^{0}\gamma91234​Pa∗→91234​Pa+00​γ.
  • Across both stages the mass number never changed, the atomic number changed only once, and only the beta stage produced a new element.

Neutron emission

Definition

Neutron emission

Neutron emission is the release of a neutron from an unstable nucleus.

  1. A neutron released from a nucleus is written 01n_{0}^{1}\text{n}01​n, giving it a mass number of 111 and a charge of 000.
  2. The nucleus therefore loses one nucleon, so its mass number falls by 111.
  3. The neutron carries no charge, so the proton count is untouched and the atomic number stays the same.
  4. Since the atomic number is unchanged, the daughter is the same element as the parent, but a different isotope of it, with one fewer neutron.
  5. In general form, neutron emission is ZAX→ZA−1X+01n_{Z}^{A}\text{X} \rightarrow _{Z}^{A-1}\text{X} + _{0}^{1}\text{n}ZA​X→ZA−1​X+01​n, so oxygen-17 emitting a neutron gives 817O→816O+01n_{8}^{17}\text{O} \rightarrow _{8}^{16}\text{O} + _{0}^{1}\text{n}817​O→816​O+01​n.
  6. Neutron emission is the one decay in this set that changes the isotope without changing the element, which makes it easy to tell apart from a beta decay in an equation.
Exam technique

Working out the effect of a decay

  • Write the emitted particle in the form chargemass numbersymbol_{\text{charge}}^{\text{mass number}}\text{symbol}chargemass number​symbol first, then subtract those two numbers from the parent.
  • Subtract the numbers separately: the top numbers form one total and the bottom numbers form another, and they are never added together.
  • Use the atomic number, not the mass number, to decide whether the element has changed, because only the proton count identifies an element.
  • Give both numbers in your answer when a question asks for the effect on the nucleus, since one mark is usually for each.
  • Say stays the same rather than leaving a number out, because a blank is not credited as an unchanged value.
  • For a gamma question, name the energy change: the nucleus loses energy, and neither the mass number nor the atomic number changes.
  • Keep to the word nucleus when describing where the emission comes from, because atom and particle are not credited in place of it.
Common Mistake
  • Do not let gamma emission change the element, because a gamma ray removes energy and carries away neither protons nor neutrons.
  • Do not reduce the mass number in a beta decay, because a nucleon changes type but no nucleon leaves the nucleus.
  • Do not reduce the atomic number in neutron emission, since a neutron carries no charge and the proton count is unaffected.
  • Do not apply two decays at once in a chain, because each step changes the numbers separately and must be worked through in order.
  • Do not treat an excited nucleus as a different element from the nucleus it settles into, because rearrangement changes only its energy.
  • Do not assume every decay produces a new element, because gamma emission and neutron emission both leave the atomic number unchanged.
Self review
  • State the change in mass number and atomic number when a nucleus emits an alpha particle.
  • Why does the mass number stay the same in a beta-minus decay?
  • A nucleus emits a positron. What happens to its atomic number?
  • Explain why gamma emission does not change the element.
  • What happens to the mass number and the atomic number when a neutron is emitted?
  • Which of these decays leaves the nucleus as a different isotope of the same element?
  • Write the general equation for the alpha decay of ZAX_{Z}^{A}\text{X}ZA​X.

6.3.3 Balancing nuclear equations

What a nuclear equation records

Definition

Nuclear equation

A nuclear equation is a way of writing a nuclear change that shows the mass number and the charge of every nucleus and particle taking part.

Definition

Mass number

The total number of protons and neutrons in the nucleus of an atom.

Definition

Charge number

The charge number is the number written at the bottom left of a symbol in a nuclear equation, giving the charge of that nucleus or particle relative to the charge of one proton.

  1. Every nucleus and particle in a nuclear equation is written as ZAX_{Z}^{A}\text{X}ZA​X, with the mass number AAA on top and the charge number ZZZ below.
  2. For a nucleus the charge number is simply the proton number, because each proton carries one unit of positive charge and the neutrons carry none.
  3. For an emitted particle the charge number is that particle's own charge, so a beta particle takes Z=−1Z = -1Z=−1 and a gamma ray takes Z=0Z = 0Z=0.
  4. Two quantities are conserved across the arrow, which is what makes the equation solvable.
    1. The total mass number on the left equals the total mass number on the right, because nucleons are neither created nor destroyed.
    2. The total charge on the left equals the total charge on the right, because charge cannot be created or destroyed.
  5. Balancing a nuclear equation therefore means finding the missing numbers that make both of those totals agree.
  6. The two totals are kept completely separate: the top numbers are added only to other top numbers, and the bottom numbers only to other bottom numbers.
  7. This is not the same task as balancing a chemical equation, where whole atoms are counted and no atom ever changes into another element.

The symbols you need

Definition

Atomic number

The number of protons in the nucleus of an atom.

  1. An alpha particle is 24α_{2}^{4}\alpha24​α, sometimes written 24He_{2}^{4}\text{He}24​He, because it is two protons and two neutrons.
  2. A beta-minus particle is −10e_{-1}^{0}\text{e}−10​e, sometimes written −10β_{-1}^{0}\beta−10​β, because it is an electron with no nucleons and one unit of negative charge.
  3. A positron is +10e_{+1}^{0}\text{e}+10​e, sometimes written +10β_{+1}^{0}\beta+10​β, because it has no nucleons and one unit of positive charge.
  4. A gamma ray is 00γ_{0}^{0}\gamma00​γ, because it carries energy but neither nucleons nor charge.
  5. A neutron is 01n_{0}^{1}\text{n}01​n, because it is one nucleon with no charge.
  6. A proton is 11p_{1}^{1}\text{p}11​p, because it is one nucleon carrying one unit of positive charge.
  7. Once these six symbols are known, every balancing task reduces to two subtractions.

The method for balancing

  1. Write the equation out with the unknown labelled, using AAA for the missing top number and ZZZ for the missing bottom number.
  2. Add the top numbers on the left, then add the top numbers on the right.
  3. Set those two totals equal to each other and solve for the missing mass number.
  4. Add the bottom numbers on the left, then add the bottom numbers on the right, remembering that a negative charge number reduces the total.
  5. Set those two totals equal to each other and solve for the missing charge number.
  6. Use the data supplied with the question, such as a section of the periodic table, to turn the charge number into an element symbol if a symbol is wanted.
  7. Substitute both answers back in and check that each total still agrees, since this catches a sign slip in seconds.
Example

Balancing an alpha decay

  • The equation to balance is 92238U→ZATh+24α_{92}^{238}\text{U} \rightarrow _{Z}^{A}\text{Th} + _{2}^{4}\alpha92238​U→ZA​Th+24​α.
  • Top numbers give 238=A+4238 = A + 4238=A+4.
  • Rearranging gives A=238−4=234A = 238 - 4 = 234A=238−4=234.
  • Bottom numbers give 92=Z+292 = Z + 292=Z+2.
  • Rearranging gives Z=92−2=90Z = 92 - 2 = 90Z=92−2=90.
  • The balanced equation is 92238U→90234Th+24α_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _{2}^{4}\alpha92238​U→90234​Th+24​α.
  • Checking gives 238=234+4238 = 234 + 4238=234+4 and 92=90+292 = 90 + 292=90+2, so both totals agree.
Example

Finding the daughter after beta-minus emission

  • The equation to balance is 1940K→ZAX+−10e_{19}^{40}\text{K} \rightarrow _{Z}^{A}\text{X} + _{-1}^{0}\text{e}1940​K→ZA​X+−10​e.
  • Top numbers give 40=A+040 = A + 040=A+0, so A=40A = 40A=40.
  • Bottom numbers give 19=Z+(−1)19 = Z + (-1)19=Z+(−1), which rearranges to Z=19+1=20Z = 19 + 1 = 20Z=19+1=20.
  • The negative charge number on the electron is what makes the daughter's charge number larger than the parent's.
  • An element with 202020 protons is calcium, so the daughter is 2040Ca_{20}^{40}\text{Ca}2040​Ca.
  • The balanced equation is 1940K→2040Ca+−10e_{19}^{40}\text{K} \rightarrow _{20}^{40}\text{Ca} + _{-1}^{0}\text{e}1940​K→2040​Ca+−10​e.
Example

Identifying an unknown emitted particle

  • The equation to balance is 1327Al→1227Mg+ZAX_{13}^{27}\text{Al} \rightarrow _{12}^{27}\text{Mg} + _{Z}^{A}\text{X}1327​Al→1227​Mg+ZA​X.
  • Top numbers give 27=27+A27 = 27 + A27=27+A, so A=0A = 0A=0 and the emitted particle contains no nucleons.
  • Bottom numbers give 13=12+Z13 = 12 + Z13=12+Z, so Z=1Z = 1Z=1 and the emitted particle carries one unit of positive charge.
  • A particle with mass number 000 and charge +1+1+1 is a positron, written +10e_{+1}^{0}\text{e}+10​e.
  • The balanced equation is 1327Al→1227Mg++10e_{13}^{27}\text{Al} \rightarrow _{12}^{27}\text{Mg} + _{+1}^{0}\text{e}1327​Al→1227​Mg++10​e.
  • Reading off the two numbers first, then naming the particle, is what keeps this kind of question quick.
Example

Balancing a decay chain of two steps

  • The chain to complete is 90232Th→Z1A1Ra+24α_{90}^{232}\text{Th} \rightarrow _{Z_1}^{A_1}\text{Ra} + _{2}^{4}\alpha90232​Th→Z1​A1​​Ra+24​α followed by Z1A1Ra→Z2A2Ac+−10e_{Z_1}^{A_1}\text{Ra} \rightarrow _{Z_2}^{A_2}\text{Ac} + _{-1}^{0}\text{e}Z1​A1​​Ra→Z2​A2​​Ac+−10​e.
  • For the first step, top numbers give A1=232−4=228A_1 = 232 - 4 = 228A1​=232−4=228 and bottom numbers give Z1=90−2=88Z_1 = 90 - 2 = 88Z1​=90−2=88.
  • The first daughter is therefore 88228Ra_{88}^{228}\text{Ra}88228​Ra.
  • For the second step, top numbers give A2=228−0=228A_2 = 228 - 0 = 228A2​=228−0=228.
  • Bottom numbers give Z2=88+1=89Z_2 = 88 + 1 = 89Z2​=88+1=89, because the electron carries away −1-1−1.
  • The second daughter is 89228Ac_{89}^{228}\text{Ac}89228​Ac, so the chain runs 90232Th→88228Ra→89228Ac_{90}^{232}\text{Th} \rightarrow _{88}^{228}\text{Ra} \rightarrow _{89}^{228}\text{Ac}90232​Th→88228​Ra→89228​Ac.
  • Each step is balanced on its own, which is always safer than trying to balance a whole chain in one line.
Exam technique

Presenting a balanced equation

  • Write both totals as a short line of working, such as 238=A+4238 = A + 4238=A+4, because the marks are usually split between the mass number and the charge number.
  • Fill in every box a question gives you, including any number that is unchanged, since a blank box scores nothing.
  • Keep the minus sign on a beta particle's charge number, because dropping it sends the daughter's charge number the wrong way by two.
  • Give the element symbol only when the question or its data allow you to identify it, and give the numbers regardless.
  • Check both totals before moving on, because a single arithmetic slip loses both marks in a two-mark part.
  • If a question supplies a periodic table extract, use it rather than recalling the symbol, since the extract is there to be read.
  • For an unknown particle, state the mass number and charge first and then name it, because that order is what the working needs to show.
Common Mistake
  • Do not add a top number to a bottom number, because mass number and charge are conserved separately.
  • Do not subtract 111 from the parent's charge number for a beta-minus particle, because a charge of −1-1−1 on the right means the daughter must be +1+1+1 higher.
  • Do not give a gamma ray a mass number or a charge, since 00γ_{0}^{0}\gamma00​γ leaves both totals untouched.
  • Do not change an element symbol without changing its charge number to match, because the two must always agree.
  • Do not treat this like a chemical equation by putting a large number in front of a symbol, because nuclear equations are balanced through the top and bottom numbers only.
Self review
  • State the two quantities that must balance on both sides of a nuclear equation.
  • Give the mass number and charge number of an alpha particle, a beta-minus particle, a positron and a gamma ray.
  • Find AAA and ZZZ in 88226Ra→ZARn+24α_{88}^{226}\text{Ra} \rightarrow _{Z}^{A}\text{Rn} + _{2}^{4}\alpha88226​Ra→ZA​Rn+24​α.
  • Find AAA and ZZZ in 614C→ZAN+−10e_{6}^{14}\text{C} \rightarrow _{Z}^{A}\text{N} + _{-1}^{0}\text{e}614​C→ZA​N+−10​e.
  • A nucleus loses a particle of mass number 000 and charge −1-1−1. Which particle is it?
  • Why does the charge number of the daughter rise when an electron is emitted?

6.4.1 Activity and its decrease over time

Activity and its unit

Definition

Activity

The activity of a radioactive source is the number of nuclei that decay each second.

Definition

Becquerel

The becquerel is the unit of activity, equal to one nuclear decay per second.

Definition

Undecayed nuclei

Undecayed nuclei are the nuclei in a sample that have not yet decayed at a given moment.

  1. A radioactive sample contains an enormous number of unstable nuclei, and each one will decay at some point, but they do not all decay together.
  2. Activity counts how fast the sample as a whole is decaying, measured as the number of nuclei that break down each second.
  3. The unit of activity is the becquerel, symbol Bq\text{Bq}Bq, and 1 Bq1\ \text{Bq}1 Bq means exactly one decay per second.
  4. Because one decay per second is a tiny rate, real sources are quoted in kilobecquerels or megabecquerels, where 1 kBq=1000 Bq1\ \text{kBq} = 1000\ \text{Bq}1 kBq=1000 Bq and 1 MBq=1 000 000 Bq1\ \text{MBq} = 1\,000\,000\ \text{Bq}1 MBq=1000000 Bq.
  5. Activity depends on two things: how many undecayed nuclei are still present, and how likely each of those nuclei is to decay in the next second.
  6. The second of those is fixed for a given isotope and cannot be altered, so heating a source, dissolving it or grinding it up has no effect on its activity.
  7. The first can only fall, because every decay permanently removes one undecayed nucleus from the sample.
  8. Activity is a rate, not an amount, so a source with a high activity is not necessarily one that contains a lot of material; it may simply be decaying quickly.

Why activity falls over time

  1. At the start there is a large stock of undecayed nuclei, so a large number of them happen to decay each second and the activity is high.
  2. Every one of those decays converts an unstable nucleus into a different nucleus, so the stock of undecayed nuclei is smaller a second later.
  3. With fewer nuclei available to decay, fewer of them decay in the next second, so the activity is lower.
  4. That lower activity in turn removes nuclei more slowly, so the fall in activity itself becomes gentler as time goes on.
  5. The result is a curve that starts steep and flattens out, always heading towards zero but never quite reaching it, which is described as exponential decay.
  6. A useful way to state the mechanism is that the activity at any moment is proportional to the number of undecayed nuclei remaining, so as one falls the other falls with it.
  7. The fall is not perfectly smooth from one reading to the next, so a run of measurements has to be repeated and averaged before the trend can be trusted.

Measuring activity in the laboratory

Definition

Count rate

The count rate is the number of decays detected each second by a radiation detector.

  1. A Geiger-Muller tube connected to a counter is the instrument used to follow how a source is decaying.
  2. What the counter records is a count rate, in counts per second, and this is not the same as the activity of the source.
  3. The count rate is always lower than the activity, for three reasons.
    1. Radiation is emitted in all directions, so only the fraction aimed at the tube can be detected.
    2. Some radiation is absorbed by the air and by the window of the tube before it can be registered.
    3. The tube does not respond to every particle that enters it, and it detects alpha radiation very poorly compared with beta and gamma.
  4. The count rate is still a valid measure of how the activity is changing, because the fraction detected stays roughly constant if the detector and the source are not moved.
  5. Every reading also includes background radiation, which is present in the room whether or not the source is there.
  6. Background must therefore be measured with the source removed and subtracted from every reading, giving a corrected count rate.
  7. In symbols, corrected count rate=measured count rate−background count rate\text{corrected count rate} = \text{measured count rate} - \text{background count rate}corrected count rate=measured count rate−background count rate.
Practical

Investigating how count rate falls with time

  • Aim: to measure how the count rate from a short-lived radioactive source falls over time, and to show that the fall is exponential.
  • Apparatus: a sealed protactinium generator bottle, a Geiger-Muller tube with a counter or datalogger, a stand, boss and clamp, a stopclock, a tray to stand the bottle in, disposable gloves and long-handled tongs.
  • Variables: time is the independent variable, count rate is the dependent variable, and the distance from the source to the tube, the position of the bottle, the counting interval and the tube's operating voltage are all controlled.
  • Method, background measurement:
    • Set up the Geiger-Muller tube in the clamp with no source anywhere near it, and switch the counter on.
    • Record the total number of counts over 300 s300\ \text{s}300 s, then divide by 300300300 to obtain the background count rate in counts per second.
    • Take this reading before the source is brought out, because background must be known before it can be subtracted.
  • Method, preparing the source:
    • Wearing gloves, lift the sealed bottle with tongs and keep it at arm's length; the bottle is never opened.
    • Shake the bottle firmly for several seconds so that the short-lived protactinium passes into the upper organic layer, then stand it in the tray and allow the two layers to separate.
    • Clamp the window of the Geiger-Muller tube against the upper layer of the bottle, close to it but not touching, and note the separation so it can be kept the same.
  • Method, taking readings:
    • Start the stopclock and the counter at the same moment, and record the number of counts in each successive 10 s10\ \text{s}10 s interval.
    • Continue for at least 200 s200\ \text{s}200 s without moving the bottle or the tube, since any change of distance would change the fraction of radiation detected.
    • Convert each interval count into a count rate by dividing by the length of the interval in seconds.
    • Subtract the background count rate from every value to obtain the corrected count rate.
    • Repeat the whole run after the bottle has been shaken again, then average corresponding readings, because single readings scatter.
  • Results: the corrected count rate falls steeply at first and then more gradually, and equal time intervals each remove the same fraction of the count rate rather than the same number of counts.
  • Processing: plot corrected count rate on the vertical axis against time on the horizontal axis, draw a smooth curve of best fit rather than joining the points, and describe the shape of the fall in words.
  • Watch out: forgetting the background subtraction flattens the tail of the curve; moving the tube during the run changes the detected fraction and puts a step in the graph; too short a counting interval gives so few counts that the scatter hides the trend.
  • Safety: keep the bottle sealed at all times, handle it with gloves and tongs, hold it away from your body, stand it in a tray in case of leakage, return it to the store immediately after use, and wash your hands afterwards.
Example

Correcting a count rate for background

  • A counter records 150015001500 counts in 300 s300\ \text{s}300 s with no source present.
  • The background count rate is 1500300=5.0\dfrac{1500}{300} = 5.03001500​=5.0 counts per second.
  • With the source in place, the counter records 360360360 counts in 10 s10\ \text{s}10 s.
  • The measured count rate is 36010=36\dfrac{360}{10} = 3610360​=36 counts per second.
  • The corrected count rate is 36−5.0=3136 - 5.0 = 3136−5.0=31 counts per second.
  • Every later reading in the same experiment must have the same 5.05.05.0 counts per second taken off it.
Example

Converting between activity and decays

  • A source is labelled as having an activity of 2.4 kBq2.4\ \text{kBq}2.4 kBq.
  • Converting the unit gives 2.4 kBq=2.4×1000=2400 Bq2.4\ \text{kBq} = 2.4 \times 1000 = 2400\ \text{Bq}2.4 kBq=2.4×1000=2400 Bq.
  • Since 1 Bq1\ \text{Bq}1 Bq is one decay per second, the source undergoes 240024002400 decays every second.
  • Over one minute the number of decays is 2400×60=144 0002400 \times 60 = 144\,0002400×60=144000.
  • This figure is only a good estimate if the activity has not fallen noticeably during that minute.
Exam technique

Writing about activity and count rate

  • Use becquerel or Bq\text{Bq}Bq as the unit of activity, and counts per second for a count rate, because the two are measured and named differently.
  • Explain the fall in activity through the number of undecayed nuclei, since saying only that the source gets weaker does not earn the reasoning mark.
  • Mention background subtraction whenever you describe or improve a radioactivity measurement, because it is a standard marking point.
  • Justify repeats by referring to the scatter between readings and the need to average, rather than to accuracy, which is not credited here.
  • Say that the detector position must be kept fixed, and give the reason that the detected fraction of the emitted radiation would otherwise change.
  • Read a value from a graph by drawing construction lines on the axes, because a bare number without them can lose the method mark.
Common Mistake
  • Do not treat the count rate as the activity, because a detector picks up only a small fraction of the radiation emitted.
  • Do not leave background radiation in your readings, because it is present whether or not the source is there.
  • Do not say that activity falls because the nuclei get tired or lose energy, since it falls because there are fewer undecayed nuclei left.
  • Do not claim that cooling, sealing or dissolving a source changes its activity, because nothing done to the material affects the chance of a nucleus decaying.
  • Do not describe the activity as reaching zero at a definite time, because the curve approaches zero without ever meeting it.
Self review
  • Define activity and give its unit.
  • How many decays per second does an activity of 3.5 kBq3.5\ \text{kBq}3.5 kBq represent?
  • Explain why the activity of a source decreases over time.
  • Why is a measured count rate always smaller than the activity of the source?
  • A background count rate of 0.80.80.8 counts per second is measured. A reading with the source gives 24.524.524.5 counts per second. What is the corrected count rate?
  • Why must the detector stay in the same position throughout a run?

6.4.2 Half-life and its random nature

What half-life means

Definition

Half-life

The half-life of a radioactive isotope is the time taken for half the undecayed nuclei in a sample to decay, or for the activity of a source to fall by half.

Definition

Undecayed nuclei

Undecayed nuclei are the nuclei in a sample that have not yet decayed at a given moment.

  1. Activity falls away smoothly towards zero, so there is no moment at which a source can be said to have finished decaying, and no way to quote a simple lifetime for it.
  2. Half-life solves this by timing the loss of a fixed fraction rather than a fixed amount, and the fraction chosen is one half.
  3. There are two equivalent statements of it, and either is acceptable.
    1. The time taken for half the undecayed nuclei in a sample to decay.
    2. The time taken for the activity of a source to fall to half its starting value.
  4. The two agree because activity is proportional to the number of undecayed nuclei, so halving one halves the other.
  5. Half-life is a time, so it is quoted in seconds, hours, days or years, and never as a fraction or a rate.
  6. Half-life is a property of the isotope, so every sample of the same isotope shares it, and it cannot be altered by heating, cooling, dissolving, compressing or chemically reacting the material.
  7. A short half-life means the activity dies away quickly, and a long half-life means a lower activity that persists for a very long time.

Half-life is constant along the curve

  1. The defining property is that the same interval of time removes the same fraction of what is left, wherever on the curve you begin.
  2. So the time to fall from the full amount to a half is the same as the time to fall from a half to a quarter, and the same again from a quarter to an eighth.
  3. This is why the fall is described as exponential: the step down is always proportional to the amount currently present.
  4. It also explains the shape of the graph, which is steep at the start where there is most left to lose and shallow later where there is least.
  5. Because the fraction lost is fixed, the amount remaining never reaches zero on the graph, and the curve approaches the time axis without meeting it.
  6. A practical consequence is that half-life can be measured from any convenient pair of points on a curve, not only from the starting value, and several such measurements should agree.

exponential-decay-and-half-life-3b514283-genie.png

Why decay is random

Definition

Random process

A random process is one in which the outcome of any individual event cannot be predicted, so it is impossible to say when a particular nucleus will decay.

  1. Nothing inside a nucleus counts down to its decay, and a nucleus does not age, wear out or become more likely to decay the longer it has already survived.
  2. Every undecayed nucleus of a given isotope simply carries the same fixed chance of decaying in the next second.
  3. It is therefore impossible to predict when a particular nucleus will decay, and impossible to say which nucleus in a sample will be next.
  4. It is equally impossible to make a chosen nucleus decay sooner or later, because nothing done to the material changes that fixed chance.
  5. What can be predicted is the behaviour of a very large number of nuclei taken together, and this is exactly what half-life does.
  6. Even a very small sample holds nuclei in numbers of the order of 102010^{20}1020, so although each one is unpredictable, the fraction that decays in a given interval is almost exactly the same every time.
  7. Half-life is therefore a statistical quantity: it predicts the activity of a large sample reliably while saying nothing whatever about any individual nucleus.
  8. The randomness is still visible when the numbers involved are small, which is why a count taken over a few seconds jumps about while the same source counted over several minutes gives a steady value.
  9. Readings taken late in the decay scatter far more than early ones, because by then too few nuclei remain for the averaging to work well.
  10. The two ideas fit together rather than contradicting each other: the process is random at the level of one nucleus and predictable at the level of a whole sample.
Practical

Modelling random decay with dice

  • Aim: to model radioactive decay with dice, showing that a random process with a fixed chance per interval produces a constant half-life.
  • Apparatus: 100100100 identical dice, a large shallow tray with raised sides, a results table, graph paper and a calculator.
  • Variables: the throw number stands for time and is the independent variable, the number of dice remaining stands for the number of undecayed nuclei and is the dependent variable, and the starting number of dice, the chance of decay per throw and the shaking technique are controlled.
  • What each part of the model stands for:
    • One die stands for one undecayed nucleus, and one throw of all the dice stands for one interval of time.
    • A die landing on a six has decayed and is removed, so every die carries the same chance of 16\dfrac{1}{6}61​ of decaying in each interval.
    • Which particular die shows a six is unpredictable, which is the feature of real decay the model exists to show.
  • Method:
    • Record 100100100 as the number remaining at throw 000, before any dice have been removed.
    • Place all 100100100 dice in the tray, shake it so that every die tumbles freely, then tip them out onto a flat surface.
    • Remove every die showing a six, count the dice left and record that number against throw 111.
    • Return only the remaining dice to the tray, so each throw starts from a smaller population than the one before.
    • Repeat for at least 151515 throws, or until fewer than five dice remain, recording the number remaining after every throw.
    • Pool the whole class's results by adding the numbers remaining at each throw number, because a larger starting population smooths out the scatter.
  • Results: the number remaining falls by roughly the same fraction at each throw, so many dice are removed early and only a few at each of the later throws.
  • Processing: plot the number of dice remaining on the vertical axis against throw number on the horizontal axis, draw a smooth curve of best fit rather than joining the points, then check that the number of throws taken to fall from 100100100 to 505050, from 505050 to 252525 and from 252525 to about 121212 come out the same.
  • Expected pattern: with a chance of 16\dfrac{1}{6}61​ per throw the model gives a half-life of roughly four throws, and the same value is obtained from any part of the curve.
  • Watch out: returning removed dice destroys the model, because a decayed nucleus does not come back; one group's data is very noisy near the end and should be pooled; failing to shake the tray properly makes the throws less than random.
  • Limitations of the model: a die that survives is completely unchanged, whereas a real decay leaves a new nucleus behind, and real decay runs continuously rather than in separate throws.
  • Safety: use a tray with raised sides so that dice are not scattered onto the floor and do not become a trip hazard.
Example

What randomness looks like in a measurement

  • A long-lived source is placed a fixed distance from a detector and five successive counts are taken, each lasting 10 s10\ \text{s}10 s.
  • The five counts come out noticeably different from one another, even though nothing about the source or the apparatus has changed.
  • The variation is not a fault in the equipment; it happens because the number of nuclei that happen to decay in any 10 s10\ \text{s}10 s window is a matter of chance.
  • Counting instead for a single interval of 600 s600\ \text{s}600 s gathers so many more decays that the same chance variation becomes a much smaller share of the total.
  • The steadier value from the longer count is the reason a radioactivity experiment uses long counting intervals and averages several repeats.
Example

Why pooling the dice results helps

  • One group starting with 100100100 dice expects about 171717 sixes on the first throw, but may well remove 111111 or 232323 instead.
  • By the twelfth throw only a handful of dice are left, so removing one die or none changes the plotted point a great deal and the curve looks ragged.
  • Adding the results of thirty groups makes the population at every throw about thirty times larger.
  • The chance variation in each group partly cancels against the others, so the pooled totals sit much closer to the smooth curve.
  • The same reasoning applies to a real source: the enormous number of nuclei present is what makes its decay curve smooth.
Exam technique

Stating half-life and randomness

  • Define half-life using the words half the undecayed nuclei or half the activity, because a loose statement about the source halving is not precise enough.
  • Include the word time in the definition, since half-life is an interval of time.
  • When asked about randomness, give both halves: an individual nucleus cannot be predicted, and a large number of nuclei can be.
  • Use the phrase cannot be predicted rather than saying decay is unknown or uncertain.
  • Justify repeat readings by naming the random nature of decay and the need to average, because reliability and consistency are credited here while accuracy is not.
  • For a question on the dice model, link each rule to the thing it represents, since the marks are for the comparison rather than for the procedure.
  • If asked why the model is limited, name a specific difference such as the discrete throws, rather than saying only that it is not realistic.
Common Mistake
  • Do not define half-life as the time for a source to decay completely, because the activity approaches zero without ever reaching it.
  • Do not define half-life as the time for half the mass of the sample to disappear, because the atoms are still there as a different element.
  • Do not say that half-life gets shorter as the source decays, because it is constant and applies from any point on the curve.
  • Do not claim that half-life can be changed by anything done to the material, because nothing alters the chance of a nucleus decaying.
  • Do not take randomness to mean the overall pattern is unpredictable, because a large sample behaves very predictably.
  • Do not say a nucleus becomes more likely to decay the longer it has survived, because its chance in the next second never changes.
Self review
  • Give two equivalent definitions of half-life.
  • Explain why the time to fall from a half to a quarter equals the time to fall from the full amount to a half.
  • Why is it impossible to predict when a particular nucleus will decay?
  • Why can the activity of a large sample still be predicted?
  • Name two things that cannot change the half-life of an isotope.
  • In the dice model, what does removing a die showing a six represent?
  • State one way in which the dice model differs from real radioactive decay.

6.4.3 Half-life calculations

Counting half-lives

Definition

Half-life

The half-life of a radioactive isotope is the time taken for half the undecayed nuclei in a sample to decay, or for the activity of a source to fall by half.

Definition

Undecayed nuclei

Undecayed nuclei are the nuclei in a sample that have not yet decayed at a given moment.

  1. Every half-life calculation turns on one number: how many half-lives have passed.
  2. The number of half-lives is n=tt1/2n = \dfrac{t}{t_{1/2}}n=t1/2​t​, where ttt is the total elapsed time and t1/2t_{1/2}t1/2​ is the half-life.
  3. Both times must be in the same unit before dividing, so a time given in hours and a half-life given in minutes must be converted first.
  4. Each half-life multiplies whatever is left by 12\dfrac{1}{2}21​, so after nnn half-lives the fraction remaining is (12)n=12n\left(\dfrac{1}{2}\right)^{n} = \dfrac{1}{2^{n}}(21​)n=2n1​.
  5. It is worth knowing the first few values by sight: 12\dfrac{1}{2}21​, 14\dfrac{1}{4}41​, 18\dfrac{1}{8}81​, 116\dfrac{1}{16}161​, 132\dfrac{1}{32}321​, 164\dfrac{1}{64}641​ for n=1n = 1n=1 to 666.
  6. The same fraction applies whether the quantity being tracked is the activity, the count rate or the number of undecayed nuclei, because all three fall together in proportion.
  7. For activity, this gives A=A0×(12)nA = A_{0} \times \left(\dfrac{1}{2}\right)^{n}A=A0​×(21​)n, where A0A_{0}A0​ is the starting activity and AAA is the activity after nnn half-lives.
  8. For nuclei, the same relationship is N=N0×(12)nN = N_{0} \times \left(\dfrac{1}{2}\right)^{n}N=N0​×(21​)n.
  9. The fraction that has decayed is whatever is left over, so it is 1−12n1 - \dfrac{1}{2^{n}}1−2n1​; after 333 half-lives, 18\dfrac{1}{8}81​ remains and 78\dfrac{7}{8}87​ has decayed.
  10. Reading the question carefully matters here, because it may ask for the amount remaining, the amount decayed, or the ratio between them, and these are three different answers.

The four question types

  1. Type one: find the amount left after a given time. Work out nnn, then halve the starting value nnn times, or multiply it by 12n\dfrac{1}{2^{n}}2n1​.
  2. Type two: find the time for a given fall. Count how many halvings take the starting value to the final value, then multiply that count by the half-life, since t=n×t1/2t = n \times t_{1/2}t=n×t1/2​.
  3. Type three: find the half-life. Count the halvings between the two given values to get nnn, then divide the elapsed time by it, since t1/2=tnt_{1/2} = \dfrac{t}{n}t1/2​=nt​.
  4. Type four: express the fall as a ratio. Give the answer in the form final to initial, such as 1:81 : 81:8, or as the fraction 18\dfrac{1}{8}81​, whichever the question asks for.
  5. When nnn is not a whole number, halving repeatedly will not reach the answer exactly, and the value must be read from a graph instead.
  6. Working in whole halvings is the quickest and safest route whenever the numbers allow it, because it needs no power key on the calculator.
Example

Finding the activity after a given time

  • A source has a starting activity of 6400 Bq6400\ \text{Bq}6400 Bq and a half-life of 151515 minutes, and the activity after 111 hour is required.
  • Converting to a common unit gives t=60t = 60t=60 minutes and t1/2=15t_{1/2} = 15t1/2​=15 minutes.
  • The number of half-lives is n=6015=4n = \dfrac{60}{15} = 4n=1560​=4.
  • The fraction remaining is 124=116\dfrac{1}{2^{4}} = \dfrac{1}{16}241​=161​.
  • Substituting gives A=6400×116=400 BqA = 6400 \times \dfrac{1}{16} = 400\ \text{Bq}A=6400×161​=400 Bq.
  • Halving four times checks this: 6400→3200→1600→800→400 Bq6400 \rightarrow 3200 \rightarrow 1600 \rightarrow 800 \rightarrow 400\ \text{Bq}6400→3200→1600→800→400 Bq.
Example

Finding the half-life from two readings

  • A count rate falls from 960960960 counts per second to 606060 counts per second in 121212 hours, and the half-life is required.
  • Halving from the start gives 960→480→240→120→60960 \rightarrow 480 \rightarrow 240 \rightarrow 120 \rightarrow 60960→480→240→120→60 counts per second.
  • That is four halvings, so n=4n = 4n=4.
  • Substituting into t1/2=tnt_{1/2} = \dfrac{t}{n}t1/2​=nt​ gives t1/2=124=3t_{1/2} = \dfrac{12}{4} = 3t1/2​=412​=3 hours.
  • Checking the fraction confirms it: 60960=116=124\dfrac{60}{960} = \dfrac{1}{16} = \dfrac{1}{2^{4}}96060​=161​=241​, which needs exactly four half-lives.
Example

Finding the time for a given fall

  • An isotope has a half-life of 888 days, and the time for its activity to fall to one eighth of its starting value is required.
  • Writing the fraction as a power of a half gives 18=123\dfrac{1}{8} = \dfrac{1}{2^{3}}81​=231​, so n=3n = 3n=3.
  • Substituting into t=n×t1/2t = n \times t_{1/2}t=n×t1/2​ gives t=3×8=24t = 3 \times 8 = 24t=3×8=24 days.
  • Stated as a ratio, the activity has changed from 888 to 111, so the final to initial ratio is 1:81 : 81:8.
  • The fraction that has decayed is 1−18=781 - \dfrac{1}{8} = \dfrac{7}{8}1−81​=87​ of the original nuclei.
Example

Working with a number of nuclei

  • A sample starts with 4.8×10124.8 \times 10^{12}4.8×1012 undecayed nuclei of an isotope whose half-life is 252525 years, and the number left after a century is required.
  • The number of half-lives is n=10025=4n = \dfrac{100}{25} = 4n=25100​=4.
  • The fraction remaining is 124=116\dfrac{1}{2^{4}} = \dfrac{1}{16}241​=161​.
  • Substituting gives N=4.8×1012×116=3.0×1011N = 4.8 \times 10^{12} \times \dfrac{1}{16} = 3.0 \times 10^{11}N=4.8×1012×161​=3.0×1011.
  • The number that has decayed is 4.8×1012−3.0×1011=4.5×10124.8 \times 10^{12} - 3.0 \times 10^{11} = 4.5 \times 10^{12}4.8×1012−3.0×1011=4.5×1012.
  • Keeping the answer in standard form avoids a power of ten error, which is the commonest slip in questions of this kind.

Reading a decay curve

Definition

Decay curve

A decay curve is a graph of activity, count rate or number of undecayed nuclei against time for a radioactive source.

  1. A decay curve plots activity, count rate or number of undecayed nuclei on the vertical axis against time on the horizontal axis.
  2. Start by reading the value on the vertical axis at t=0t = 0t=0, which is the starting value A0A_{0}A0​.
  3. Halve that value, find the halved value on the vertical axis, and draw a horizontal construction line across to the curve.
  4. From that meeting point, drop a vertical construction line down to the time axis and read off the time, which is the half-life.
  5. Repeat the same steps starting from a different point on the curve, for instance from the half value down to the quarter value, and check that the two times agree.
  6. Averaging two or three such readings gives a better value than one, because the curve is drawn through scattered points.
  7. Leave the construction lines on the graph, since they show the method and are what a marker looks for.
  8. Where the vertical axis is a count rate, subtract any background count rate the question supplies before halving, otherwise the halving is applied to the wrong starting value.
  9. To read the value at a time between plotted points, go up from the time axis to the curve and across, rather than trying to calculate it.

exponential-decay-and-half-life-3b514283-genie.png

Example

Taking a half-life from a graph

  • A decay curve of count rate against time starts at 808080 counts per second at t=0t = 0t=0.
  • Half of 808080 is 404040, so a horizontal line is drawn from 404040 on the vertical axis across to the curve.
  • Dropping a vertical line from that point meets the time axis at 666 minutes, giving a first half-life value of 666 minutes.
  • For a second reading, half of 404040 is 202020, and the same construction gives a time of 121212 minutes.
  • The interval between the two readings is 12−6=612 - 6 = 612−6=6 minutes, which agrees with the first value.
  • Since both readings give 666 minutes, the half-life is 666 minutes.
Exam technique

Setting out a half-life calculation

  • Show the number of half-lives as a separate line of working, because nnn is usually worth a mark of its own.
  • Print the substitution with the numbers in it, such as A=6400×116A = 6400 \times \dfrac{1}{16}A=6400×161​, rather than only the rearranged formula.
  • Convert both times into the same unit before dividing, and state the unit you converted to.
  • Give the unit with the final answer, since a bare number can be marked down where the unit carries a mark.
  • Write out the successive halvings when the numbers are friendly, because each halving is visible working that can earn credit if the final value slips.
  • Answer the question that was asked: remaining, decayed, or a ratio are three different answers from the same value of nnn.
  • For a graph question, draw and leave the construction lines, and take two readings from different parts of the curve.
  • Keep large or small numbers in standard form throughout, because a power of ten error can cost every mark after it.
Common Mistake
  • Do not multiply the starting value by nnn or divide it by 2n2n2n, because each half-life multiplies what is left by 12\dfrac{1}{2}21​ and the effect compounds.
  • Do not assume the activity is zero once several half-lives have passed, because a fraction always remains.
  • Do not divide the elapsed time by the half-life and then treat the result as the answer, because that quotient is nnn and not the activity.
  • Do not give the fraction remaining when the question asks for the fraction that has decayed, since the two add to 111.
  • Do not read a half-life from a graph without halving the starting value first, because the interval must run between a value and its half.
  • Do not forget to subtract a stated background count rate before halving, or the starting value will be too high.
  • Do not mix units, because a half-life in minutes with a time in hours gives an answer wrong by a factor of 606060.
Self review
  • Write the equation linking the number of half-lives, the elapsed time and the half-life.
  • What fraction of the original nuclei remains after 555 half-lives?
  • A source of activity 1200 Bq1200\ \text{Bq}1200 Bq has a half-life of 202020 minutes. What is its activity after 111 hour?
  • A count rate falls from 640640640 to 404040 counts per second in 181818 hours. What is the half-life?
  • An isotope has a half-life of 555 days. How long does its activity take to fall to one sixteenth of its starting value?
  • Describe how to find a half-life from a graph of count rate against time.
  • After 444 half-lives, what fraction of the original nuclei has decayed?

6.6.1 Diagnosis and treatment using radiation

Treating tumours with radiation

Definition

Radiotherapy

Radiotherapy is the treatment of a tumour with high-energy ionising radiation in order to kill cancer cells or stop them dividing.

  1. The radiation used is high-energy gamma rays or high-energy X-rays, because only very penetrating radiation can reach a tumour buried inside the body.
  2. As the radiation passes through a cell it ionises molecules, and the damage that matters most is the damage done to the cell's DNA.
  3. A cell with badly damaged DNA cannot copy it correctly, so the cell fails when it next tries to divide and it dies.
  4. Tumour cells divide far more often than most healthy cells and are worse at repairing DNA damage, so the same dose kills a much greater share of them.
  5. Healthy cells in the path of the radiation are damaged too, which is what causes the side effects of treatment, so every method of delivering the radiation is designed to give the largest possible dose to the tumour and the smallest possible dose everywhere else.
  6. The total dose is usually split into many small fractions given on different days, which gives healthy tissue time to repair between sessions while the tumour is worn down.
  7. The radiation can be delivered from a machine outside the body or from a source placed inside it.

External radiotherapy

  1. A machine outside the patient produces a narrow beam of high-energy radiation and directs it at the tumour, so no radioactive material ever enters the body.
  2. The beam is shaped by a collimator so that its cross-section matches the outline of the tumour as closely as possible.
  3. The machine is then rotated around the patient, so the beam enters the body from many different directions during the session.
  4. Every beam direction passes through the tumour, so the tumour receives the sum of all the doses and ends up with a very large total.
  5. Any one region of healthy tissue lies in the path of only one or two of those directions, so it receives a much smaller total than the tumour does.
  6. Advantages: nothing is implanted, so no surgery is needed; the patient is not radioactive at any point and can go home straight away; deep tumours can be reached; and the direction, shape and size of the dose can be adjusted between sessions.
  7. Disadvantages: the radiation has to travel through healthy tissue on its way in and out, side effects such as tiredness, reddened skin and hair loss in the treated area are common, many sessions are needed over several weeks, and the patient must be positioned very accurately each time.

Internal radiotherapy

  1. A radioactive source is placed inside the body, either in the tumour itself or immediately next to it.
  2. The source may be a small sealed pellet or rod inserted during a short procedure, or a radioactive substance that the body itself concentrates in the target tissue.
  3. An overactive thyroid or a thyroid tumour, for example, can be treated with radioactive iodine, because the thyroid gland absorbs iodine from the blood whether or not the nuclei are stable.
  4. The radiation spreads out from the source in all directions, so its intensity falls rapidly with distance.
  5. Tissue touching the source therefore receives an extremely large dose while tissue a few centimetres further away receives very little, which is exactly the pattern the treatment wants.
  6. Advantages: the tumour gets a very high dose, healthy tissue further away is largely spared, treatment continues steadily while the source stays in place, and fewer hospital visits are needed.
  7. Disadvantages: placing the source usually needs an invasive procedure, staff have to handle a radioactive source, the patient is radioactive while the source is inside them and so must limit close contact with visitors, and the method only suits tumours that can be reached or that take up a particular substance.
Key Idea
  • Both treatments use ionising radiation to damage the DNA of cancer cells so they cannot divide.
  • The difference is where the source is: outside the body and aimed in, or inside the body and already at the target.
  • External treatment spares healthy tissue by crossing beams, internal treatment spares it by being close.

Tracers in diagnosis

Definition

Radioactive tracer

A radioactive tracer is a small quantity of a radioactive substance added to a system so that its position or movement can be followed using a radiation detector.

  1. A tracer is swallowed, injected or inhaled, depending on which organ is being investigated, and is carried around the body by the blood.
  2. The radioactive atoms are attached to a chemical the target tissue takes up naturally, so the tracer collects where the doctor wants to look.
  3. A gamma camera outside the patient detects the radiation coming out and builds a picture of where the tracer has gone.
  4. A medical tracer must be a gamma emitter, because gamma radiation is the only kind penetrating enough to escape the body and reach the camera; alpha and beta radiation would be absorbed by the surrounding tissue and would only add to the patient's dose.
  5. It must also have a short half-life, long enough for the scan to be completed but short enough that the activity inside the patient falls away within hours rather than years.
  6. Technetium-99m is the isotope used most often, because it emits gamma radiation only and has a half-life of about 6 hours6\ \text{hours}6 hours.
  7. The doctor is looking for tissue that takes up more or less tracer than expected, since that pattern shows how the organ is working rather than just what shape it is.
  8. Tracers are used to follow blood flow, to check how well a kidney or a thyroid gland is functioning, to find blockages in the circulation or the digestive system, and to locate secondary tumours in bone.

How a PET scanner works

Definition

PET scan

A PET scan is a medical image built from the gamma rays produced when positrons emitted by a tracer inside the body meet electrons in nearby tissue.

Definition

Positron

A positron is the antiparticle of the electron, with the same mass as an electron but a positive charge.

  1. The patient is injected with a tracer whose nuclei emit positrons rather than gamma rays, most often fluorine-18 attached to a molecule that behaves like glucose.
  2. Tissue that is working hardest uses the most glucose, so the most active tissue, including many tumours, takes up the most tracer.
  3. Each emitted positron travels only a very short distance through tissue before it meets an electron, of which there are enormous numbers in every atom around it.
  4. A positron and an electron are a particle and its antiparticle, so when they meet they annihilate: both are destroyed and their mass is converted into energy.
  5. That energy leaves as two gamma rays travelling in opposite directions, which is required because the pair had almost no momentum before they met.
  6. A ring of detectors surrounds the patient, and the electronics look for two gamma rays arriving on opposite sides of the ring at almost the same instant.
  7. Each such pair of detections means an annihilation happened somewhere along the straight line joining the two detectors.
  8. A computer collects millions of these lines and finds the regions where they cross most often, and those regions are where the tracer is concentrated.
  9. The result is a three-dimensional image showing how active each part of the body is, which is information that an X-ray image of structure alone cannot give.
  10. PET scans are used to find tumours, to check whether a tumour is still active after treatment, and to study how the brain and the heart are working.

Making PET isotopes nearby

  1. The isotopes that emit positrons have very short half-lives: about 110 minutes110\ \text{minutes}110 minutes for fluorine-18 and only about 20 minutes20\ \text{minutes}20 minutes for carbon-11.
  2. A short half-life is chosen deliberately, because it means the activity inside the patient falls away within a few hours and the dose is kept small.
  3. The same property applies before the scan as well, so the activity is falling steeply from the moment the isotope is made.
  4. A long journey would use up several half-lives, and once too few nuclei are left the number of annihilations per second is too small to build a clear image.
  5. Giving the patient a much larger amount to compensate is not an option, because that would raise the dose and undo the reason for using a short-lived isotope.
  6. The isotopes are therefore made in a cyclotron on the hospital site or at a centre close to it, attached to the tracer molecule immediately, and injected within a few hours.
  7. This is why PET scanning is expensive and why it is available at far fewer hospitals than ordinary X-ray imaging.
Example

Activity lost in transport

  • A batch of fluorine-18, half-life 110 minutes110\ \text{minutes}110 minutes, is made at a cyclotron with an activity of 800 MBq800\ \text{MBq}800 MBq.
  • Preparing it and driving it to a distant hospital takes 220 minutes220\ \text{minutes}220 minutes.
  • The number of half-lives is n=tT1/2=220110=2n=\dfrac{t}{T_{1/2}}=\dfrac{220}{110}=2n=T1/2​t​=110220​=2.
  • Using A=A0(12)nA=A_{0}\left(\dfrac{1}{2}\right)^{n}A=A0​(21​)n gives A=800×(12)2=200 MBqA=800\times\left(\dfrac{1}{2}\right)^{2}=200\ \text{MBq}A=800×(21​)2=200 MBq.
  • Three quarters of the activity has been lost before the tracer even reaches the patient.
  • Carbon-11, with a half-life of 20 minutes20\ \text{minutes}20 minutes, would pass through eleven half-lives in the same journey and arrive with less than one thousandth of its activity, which is why it can only be used where it is made.
Exam technique

Comparing and explaining

  • For a compare question on treatment, give at least one similarity and one difference, and make the difference about where the source is rather than about which is better.
  • When you justify a property of a tracer, link it to a consequence: gamma so the radiation can leave the body, short half-life so the activity soon falls and the dose stays low.
  • For PET, describe the whole chain in order, because each stage is a separate marking point: tracer injected, positron emitted, positron meets electron, annihilation, two gamma rays in opposite directions, detected by a ring, computer builds the image.
  • For the question about producing isotopes nearby, three linked steps are needed: short half-life, activity falls quickly during transport, too little activity left for a clear scan.
  • Where numbers are given, work in half-lives first, since n=tT1/2n=\dfrac{t}{T_{1/2}}n=T1/2​t​ is usually worth a mark of its own.
Common Mistake
  • Do not say a PET tracer emits gamma rays directly; it emits positrons, and the gamma rays are produced when a positron meets an electron.
  • Do not say the two gamma rays travel in the same direction; they travel in opposite directions, which is what allows the position to be worked out.
  • Do not choose an alpha or beta emitter as a tracer; the radiation would never get out of the patient to be detected.
  • Do not say a short half-life is chosen only to protect the patient; it is also the reason the isotope cannot be transported far.
  • Do not write that external radiotherapy leaves the patient radioactive; only internal treatment puts a source inside the body.
  • Do not claim radiotherapy affects only cancer cells; healthy cells in the beam are damaged as well, which is why the dose is spread over many sessions.
Self review
  • Explain why cancer cells are killed by a dose of radiation that healthy cells survive.
  • Explain how rotating the beam around the patient protects healthy tissue during external radiotherapy.
  • Give one similarity and two differences between internal and external treatment of a tumour.
  • State the two properties a medical tracer must have and explain why each is needed.
  • Describe, in order, what happens between a positron being emitted and a PET image appearing.
  • Explain why isotopes used in PET scanners have to be produced close to the scanner.

Recap questions

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In Rutherford's gold foil experiment, most alpha particles passed straight through the foil. What does this show about atoms?

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Rutherford scattering experiment diagram

In the early 1900s, scientists thought atoms looked like "plum puddings": spheres of positive charge with tiny negative electrons scattered inside them.

This idea changed after Rutherford's alpha scattering experiment. When positive alpha particles were fired at thin gold foil, most went straight through, but a few rebounded at sharp angles.

This proved that an atom is mostly empty space. It contains a tiny, dense, positively charged nucleus at its center, with electrons orbiting far around it.

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Radioactive tracers can be used to monitor how well a patient's lungs are working.

A radioactive gas is breathed in by the patient. The isotope emits radiation.

As the gas circulates through the lungs, this radiation is detected outside the chest by a scanner.

What type of radiation is able to pass through the body tissue to reach the scanner?

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How is the number of neutrons in a nucleus calculated?

6.3 Radioactive decay, half-life and uses Revision Guide

  1. GCSE
  2. /Physics
  3. /6.3 Radioactive decay, half-life and uses

Revision notes for Edexcel GCSE Physics 6.3 Radioactive decay, half-life and uses. Open the guide for explanations and worked examples. Written against the Edexcel GCSE Physics (1PH0) specification, so the content matches what's examinable rather than general Physics background.