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.

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.
When a nucleus holds too many neutrons or too many protons for that balance to hold, it is unstable and it will eventually decay.
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.
Instead, one nucleon changes identity inside the nucleus, and a small light particle is created and thrown out to carry away the charge.
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 $β−\beta^-$ 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 $β+\beta^+$ decay.
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.

Inside a neutron-rich nucleus, a single neutron changes into a proton and an electron, and the electron is immediately expelled from the nucleus.
In words, the change to the nucleon is $neutron→proton+electron\text{neutron} \rightarrow \text{proton} + \text{electron}$ .
Using particle symbols, the same change is $n→p+e−\text{n} \rightarrow \text{p} + \text{e}^-$ .
Written with mass numbers and charges, so that both quantities can be seen to balance, the change is $01n→11p+−10e_{0}^{1}\text{n} \rightarrow _{1}^{1}\text{p} + _{-1}^{0}\text{e}$ .
The emitted electron is the beta-minus particle, and in a nuclear equation it is written as $−10e_{-1}^{0}\text{e}$ or $−10β_{-1}^{0}\beta$ .
Both symbols carry the same information: a mass number of $0$ , because an electron contains no protons or neutrons, and a charge of $- 1$ , because it carries one electron charge.
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.
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 $614C_{6}^{14}\text{C}$ , so it has $6$ protons and $14 - 6 = 8$ neutrons.
Eight neutrons to six protons is a neutron-rich mixture for such a light nucleus, so this nucleus decays by $β−\beta^-$ emission.
One neutron becomes a proton, so the proton count rises from $6$ to $7$ while the total nucleon count stays at $14$ .
An element with $7$ protons is nitrogen, so the daughter nucleus is $714N_{7}^{14}\text{N}$ .
The decay is written $614C→714N+−10e_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e}$ , 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.

Inside a proton-rich nucleus, a single proton changes into a neutron and a positron, and the positron is immediately expelled from the nucleus.
In words, the change to the nucleon is $proton→neutron+positron\text{proton} \rightarrow \text{neutron} + \text{positron}$ .
Using particle symbols, the same change is $p→n+e+\text{p} \rightarrow \text{n} + \text{e}^+$ .
Written with mass numbers and charges, the change is $11p→01n++10e_{1}^{1}\text{p} \rightarrow _{0}^{1}\text{n} + _{+1}^{0}\text{e}$ .
The emitted positron is written $+10e_{+1}^{0}\text{e}$ or $+10β_{+1}^{0}\beta$ , with a mass number of $0$ and a charge of $+ 1$ .
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.
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 $918F_{9}^{18}\text{F}$ , so it has $9$ protons and $18 - 9 = 9$ neutrons.
Nine protons to nine neutrons leaves this nucleus proton-rich compared with the stable isotope fluorine-19, so it decays by $β+\beta^+$ emission.
One proton becomes a neutron, so the proton count falls from $9$ to $8$ while the total nucleon count stays at $18$ .
An element with $8$ protons is oxygen, so the daughter nucleus is $818O_{8}^{18}\text{O}$ .
The decay is written $918F→818O++10e_{9}^{18}\text{F} \rightarrow _{8}^{18}\text{O} + _{+1}^{0}\text{e}$ , 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

Both decays start with an unstable nucleus, change one nucleon into the other kind of nucleon, and emit one light particle from the nucleus.
The differences all follow from which nucleon changes.
1. In $β−\beta^-$ decay a neutron is the nucleon that changes, so the neutron count falls by one and the proton count rises by one.
2. In $β+\beta^+$ decay a proton is the nucleon that changes, so the proton count falls by one and the neutron count rises by one.
The emitted particles differ only in the sign of their charge, which is what the minus and plus in the two names record.
Neither decay ejects a nucleon from the nucleus, which is what separates beta decay from alpha decay and from neutron emission.
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 $β−\beta^-$ decay as a neutron changing into a proton with an electron emitted, and $β+\beta^+$ 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, $−10e_{-1}^{0}\text{e}$ , and plus goes with the positron, $+10e_{+1}^{0}\text{e}$ .
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 $1$ , since its mass number is $0$ 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 $β−\beta^-$ decay, and which particle is emitted?
What happens to a proton during $β+\beta^+$ 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 $β+\beta^+$ decay.
Explain why the electron emitted in $β−\beta^-$ 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.

A nucleus is written in the form $ZAX_{Z}^{A}\text{X}$ , where $X\text{X}$ is the chemical symbol, $A$ is the mass number and $Z$ is the atomic number.
The atomic number is the count of protons, and it alone decides which element the nucleus belongs to, so any decay that changes $Z$ changes the element.
The mass number is the count of protons plus neutrons, so the neutron count of a nucleus is always $A - Z$ .
Every radioactive decay removes something from the nucleus or converts one nucleon into another, so every decay leaves $A$ and $Z$ changed in a way fixed by what was emitted.
The nucleus that is left behind after a decay is the daughter nucleus, and the nucleus that decayed is called the parent nucleus.
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.

An alpha particle is a tightly bound group of two protons and two neutrons, identical to a helium nucleus, and it is written $24α_{2}^{4}\alpha$ or $24He_{2}^{4}\text{He}$ .
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.
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.
The alpha particle carries away a mass number of $4$ , so the mass number of the nucleus falls by $4$ .
It also carries away a charge of $+ 2$ , so the atomic number falls by $2$ .
Since the atomic number has changed, the daughter nucleus is a different element, two places back in the periodic table.
In general form, alpha decay is $ZAX→Z−2A−4Y+24α_{Z}^{A}\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\alpha$ .

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 $88226Ra_{88}^{226}\text{Ra}$ , so it has $88$ protons and $226 - 88 = 138$ neutrons.
The mass number of the daughter is $226 - 4 = 222$ .
The atomic number of the daughter is $88 - 2 = 86$ .
The element with $86$ protons is radon, so the daughter nucleus is $86222Rn_{86}^{222}\text{Rn}$ .
The complete equation is $88226Ra→86222Rn+24α_{88}^{226}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\alpha$ .
The neutron count has fallen from $138$ to $222 - 86 = 136$ , which confirms that two protons and two neutrons left together.

Beta-minus and beta-plus decay

In beta-minus decay a neutron becomes a proton and an electron is emitted, and the electron is written $−10e_{-1}^{0}\text{e}$ .
Because the electron has a mass number of $0$ , the total nucleon count is unchanged, so the mass number stays the same.
Because a neutron has been replaced by a proton, the atomic number rises by $1$ , and the daughter is the next element up.
In general form, beta-minus decay is $ZAX→Z+1AY+−10e_{Z}^{A}\text{X} \rightarrow _{Z+1}^{A}\text{Y} + _{-1}^{0}\text{e}$ , as in $614C→714N+−10e_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e}$ .
In beta-plus decay a proton becomes a neutron and a positron is emitted, and the positron is written $+10e_{+1}^{0}\text{e}$ .
The mass number again stays the same, for the same reason, but the atomic number falls by $1$ , so the daughter is the previous element.
In general form, beta-plus decay is $ZAX→Z−1AY++10e_{Z}^{A}\text{X} \rightarrow _{Z-1}^{A}\text{Y} + _{+1}^{0}\text{e}$ , as in $1122Na→1022Ne++10e_{11}^{22}\text{Na} \rightarrow _{10}^{22}\text{Ne} + _{+1}^{0}\text{e}$ .
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.

Gamma radiation is not a particle at all; it is a very high frequency electromagnetic wave carrying energy away from the nucleus.
A gamma ray has no mass number and no charge, so it is written $00γ_{0}^{0}\gamma$ .
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.
What does change is the energy of the nucleus, which is left lower than it was before.
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.
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.
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.
In general form, gamma emission is $ZAX∗→ZAX+00γ_{Z}^{A}\text{X}^{*} \rightarrow _{Z}^{A}\text{X} + _{0}^{0}\gamma$ , 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 $90234Th_{90}^{234}\text{Th}$ , is neutron-rich and decays by beta-minus emission.
The mass number stays at $234$ and the atomic number rises from $90$ to $91$ , giving protactinium-234.
The equation for the first stage is $90234Th→91234Pa+−10e_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}\text{e}$ .
The protactinium nucleus is left excited, so its nucleons rearrange and it emits a gamma ray.
The equation for the second stage is $91234Pa∗→91234Pa+00γ_{91}^{234}\text{Pa}^{*} \rightarrow _{91}^{234}\text{Pa} + _{0}^{0}\gamma$ .
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.

A neutron released from a nucleus is written $01n_{0}^{1}\text{n}$ , giving it a mass number of $1$ and a charge of $0$ .
The nucleus therefore loses one nucleon, so its mass number falls by $1$ .
The neutron carries no charge, so the proton count is untouched and the atomic number stays the same.
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.
In general form, neutron emission is $ZAX→ZA−1X+01n_{Z}^{A}\text{X} \rightarrow _{Z}^{A-1}\text{X} + _{0}^{1}\text{n}$ , so oxygen-17 emitting a neutron gives $817O→816O+01n_{8}^{17}\text{O} \rightarrow _{8}^{16}\text{O} + _{0}^{1}\text{n}$ .
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 $numbersymbol_{\text{charge}}^{\text{mass number}}\text{symbol}$ 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 $ZAX_{Z}^{A}\text{X}$ .

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.

Every nucleus and particle in a nuclear equation is written as $ZAX_{Z}^{A}\text{X}$ , with the mass number $A$ on top and the charge number $Z$ below.
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.
For an emitted particle the charge number is that particle's own charge, so a beta particle takes $Z = - 1$ and a gamma ray takes $Z = 0$ .
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.
Balancing a nuclear equation therefore means finding the missing numbers that make both of those totals agree.
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.
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.

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

The method for balancing

Write the equation out with the unknown labelled, using $A$ for the missing top number and $Z$ for the missing bottom number.
Add the top numbers on the left, then add the top numbers on the right.
Set those two totals equal to each other and solve for the missing mass number.
Add the bottom numbers on the left, then add the bottom numbers on the right, remembering that a negative charge number reduces the total.
Set those two totals equal to each other and solve for the missing charge number.
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.
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 $92238U→ZATh+24α_{92}^{238}\text{U} \rightarrow _{Z}^{A}\text{Th} + _{2}^{4}\alpha$ .
Top numbers give $238 = A + 4$ .
Rearranging gives $A = 238 - 4 = 234$ .
Bottom numbers give $92 = Z + 2$ .
Rearranging gives $Z = 92 - 2 = 90$ .
The balanced equation is $92238U→90234Th+24α_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _{2}^{4}\alpha$ .
Checking gives $238 = 234 + 4$ and $92 = 90 + 2$ , so both totals agree.

Example

Finding the daughter after beta-minus emission

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

Example

Identifying an unknown emitted particle

The equation to balance is $1327Al→1227Mg+ZAX_{13}^{27}\text{Al} \rightarrow _{12}^{27}\text{Mg} + _{Z}^{A}\text{X}$ .
Top numbers give $27 = 27 + A$ , so $A = 0$ and the emitted particle contains no nucleons.
Bottom numbers give $13 = 12 + Z$ , so $Z = 1$ and the emitted particle carries one unit of positive charge.
A particle with mass number $0$ and charge $+ 1$ is a positron, written $+10e_{+1}^{0}\text{e}$ .
The balanced equation is $1327Al→1227Mg++10e_{13}^{27}\text{Al} \rightarrow _{12}^{27}\text{Mg} + _{+1}^{0}\text{e}$ .
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 $90232Th→Z1A1Ra+24α_{90}^{232}\text{Th} \rightarrow _{Z_1}^{A_1}\text{Ra} + _{2}^{4}\alpha$ followed by $Z1A1Ra→Z2A2Ac+−10e_{Z_1}^{A_1}\text{Ra} \rightarrow _{Z_2}^{A_2}\text{Ac} + _{-1}^{0}\text{e}$ .
For the first step, top numbers give $A_1 = 232 - 4 = 228$ and bottom numbers give $Z_1 = 90 - 2 = 88$ .
The first daughter is therefore $88228Ra_{88}^{228}\text{Ra}$ .
For the second step, top numbers give $A_2 = 228 - 0 = 228$ .
Bottom numbers give $Z_2 = 88 + 1 = 89$ , because the electron carries away $- 1$ .
The second daughter is $89228Ac_{89}^{228}\text{Ac}$ , so the chain runs $90232Th→88228Ra→89228Ac_{90}^{232}\text{Th} \rightarrow _{88}^{228}\text{Ra} \rightarrow _{89}^{228}\text{Ac}$ .
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 $238 = A + 4$ , 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 $1$ from the parent's charge number for a beta-minus particle, because a charge of $- 1$ on the right means the daughter must be $+ 1$ higher.
Do not give a gamma ray a mass number or a charge, since $00γ_{0}^{0}\gamma$ 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 $A$ and $Z$ in $88226Ra→ZARn+24α_{88}^{226}\text{Ra} \rightarrow _{Z}^{A}\text{Rn} + _{2}^{4}\alpha$ .
Find $A$ and $Z$ in $614C→ZAN+−10e_{6}^{14}\text{C} \rightarrow _{Z}^{A}\text{N} + _{-1}^{0}\text{e}$ .
A nucleus loses a particle of mass number $0$ and charge $- 1$ . Which particle is it?
Why does the charge number of the daughter rise when an electron is emitted?

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.

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.
When a nucleus holds too many neutrons or too many protons for that balance to hold, it is unstable and it will eventually decay.
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.
Instead, one nucleon changes identity inside the nucleus, and a small light particle is created and thrown out to carry away the charge.
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 $β−\beta^-$ 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 $β+\beta^+$ decay.
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.

Inside a neutron-rich nucleus, a single neutron changes into a proton and an electron, and the electron is immediately expelled from the nucleus.
In words, the change to the nucleon is $neutron→proton+electron\text{neutron} \rightarrow \text{proton} + \text{electron}$ .
Using particle symbols, the same change is $n→p+e−\text{n} \rightarrow \text{p} + \text{e}^-$ .
Written with mass numbers and charges, so that both quantities can be seen to balance, the change is $01n→11p+−10e_{0}^{1}\text{n} \rightarrow _{1}^{1}\text{p} + _{-1}^{0}\text{e}$ .
The emitted electron is the beta-minus particle, and in a nuclear equation it is written as $−10e_{-1}^{0}\text{e}$ or $−10β_{-1}^{0}\beta$ .
Both symbols carry the same information: a mass number of $0$ , because an electron contains no protons or neutrons, and a charge of $- 1$ , because it carries one electron charge.
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.
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.

Example

Carbon-14 decaying by beta-minus emission

Carbon-14 is written $614C_{6}^{14}\text{C}$ , so it has $6$ protons and $14 - 6 = 8$ neutrons.
Eight neutrons to six protons is a neutron-rich mixture for such a light nucleus, so this nucleus decays by $β−\beta^-$ emission.
One neutron becomes a proton, so the proton count rises from $6$ to $7$ while the total nucleon count stays at $14$ .
An element with $7$ protons is nitrogen, so the daughter nucleus is $714N_{7}^{14}\text{N}$ .
The decay is written $614C→714N+−10e_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e}$ , 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.

Inside a proton-rich nucleus, a single proton changes into a neutron and a positron, and the positron is immediately expelled from the nucleus.
In words, the change to the nucleon is $proton→neutron+positron\text{proton} \rightarrow \text{neutron} + \text{positron}$ .
Using particle symbols, the same change is $p→n+e+\text{p} \rightarrow \text{n} + \text{e}^+$ .
Written with mass numbers and charges, the change is $11p→01n++10e_{1}^{1}\text{p} \rightarrow _{0}^{1}\text{n} + _{+1}^{0}\text{e}$ .
The emitted positron is written $+10e_{+1}^{0}\text{e}$ or $+10β_{+1}^{0}\beta$ , with a mass number of $0$ and a charge of $+ 1$ .
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.
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 $918F_{9}^{18}\text{F}$ , so it has $9$ protons and $18 - 9 = 9$ neutrons.
Nine protons to nine neutrons leaves this nucleus proton-rich compared with the stable isotope fluorine-19, so it decays by $β+\beta^+$ emission.
One proton becomes a neutron, so the proton count falls from $9$ to $8$ while the total nucleon count stays at $18$ .
An element with $8$ protons is oxygen, so the daughter nucleus is $818O_{8}^{18}\text{O}$ .
The decay is written $918F→818O++10e_{9}^{18}\text{F} \rightarrow _{8}^{18}\text{O} + _{+1}^{0}\text{e}$ , 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

Both decays start with an unstable nucleus, change one nucleon into the other kind of nucleon, and emit one light particle from the nucleus.
The differences all follow from which nucleon changes.
1. In $β−\beta^-$ decay a neutron is the nucleon that changes, so the neutron count falls by one and the proton count rises by one.
2. In $β+\beta^+$ decay a proton is the nucleon that changes, so the proton count falls by one and the neutron count rises by one.
The emitted particles differ only in the sign of their charge, which is what the minus and plus in the two names record.
Neither decay ejects a nucleon from the nucleus, which is what separates beta decay from alpha decay and from neutron emission.
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 $β−\beta^-$ decay as a neutron changing into a proton with an electron emitted, and $β+\beta^+$ 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, $−10e_{-1}^{0}\text{e}$ , and plus goes with the positron, $+10e_{+1}^{0}\text{e}$ .
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 $1$ , since its mass number is $0$ 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 $β−\beta^-$ decay, and which particle is emitted?
What happens to a proton during $β+\beta^+$ 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 $β+\beta^+$ decay.
Explain why the electron emitted in $β−\beta^-$ 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.

A nucleus is written in the form $ZAX_{Z}^{A}\text{X}$ , where $X\text{X}$ is the chemical symbol, $A$ is the mass number and $Z$ is the atomic number.
The atomic number is the count of protons, and it alone decides which element the nucleus belongs to, so any decay that changes $Z$ changes the element.
The mass number is the count of protons plus neutrons, so the neutron count of a nucleus is always $A - Z$ .
Every radioactive decay removes something from the nucleus or converts one nucleon into another, so every decay leaves $A$ and $Z$ changed in a way fixed by what was emitted.
The nucleus that is left behind after a decay is the daughter nucleus, and the nucleus that decayed is called the parent nucleus.
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.

An alpha particle is a tightly bound group of two protons and two neutrons, identical to a helium nucleus, and it is written $24α_{2}^{4}\alpha$ or $24He_{2}^{4}\text{He}$ .
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.
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.
The alpha particle carries away a mass number of $4$ , so the mass number of the nucleus falls by $4$ .
It also carries away a charge of $+ 2$ , so the atomic number falls by $2$ .
Since the atomic number has changed, the daughter nucleus is a different element, two places back in the periodic table.
In general form, alpha decay is $ZAX→Z−2A−4Y+24α_{Z}^{A}\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\alpha$ .

Example

Finding the daughter of an alpha emitter

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

Beta-minus and beta-plus decay

In beta-minus decay a neutron becomes a proton and an electron is emitted, and the electron is written $−10e_{-1}^{0}\text{e}$ .
Because the electron has a mass number of $0$ , the total nucleon count is unchanged, so the mass number stays the same.
Because a neutron has been replaced by a proton, the atomic number rises by $1$ , and the daughter is the next element up.
In general form, beta-minus decay is $ZAX→Z+1AY+−10e_{Z}^{A}\text{X} \rightarrow _{Z+1}^{A}\text{Y} + _{-1}^{0}\text{e}$ , as in $614C→714N+−10e_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e}$ .
In beta-plus decay a proton becomes a neutron and a positron is emitted, and the positron is written $+10e_{+1}^{0}\text{e}$ .
The mass number again stays the same, for the same reason, but the atomic number falls by $1$ , so the daughter is the previous element.
In general form, beta-plus decay is $ZAX→Z−1AY++10e_{Z}^{A}\text{X} \rightarrow _{Z-1}^{A}\text{Y} + _{+1}^{0}\text{e}$ , as in $1122Na→1022Ne++10e_{11}^{22}\text{Na} \rightarrow _{10}^{22}\text{Ne} + _{+1}^{0}\text{e}$ .
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.

Gamma radiation is not a particle at all; it is a very high frequency electromagnetic wave carrying energy away from the nucleus.
A gamma ray has no mass number and no charge, so it is written $00γ_{0}^{0}\gamma$ .
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.
What does change is the energy of the nucleus, which is left lower than it was before.
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.
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.
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.
In general form, gamma emission is $ZAX∗→ZAX+00γ_{Z}^{A}\text{X}^{*} \rightarrow _{Z}^{A}\text{X} + _{0}^{0}\gamma$ , where the asterisk marks the excited nucleus.

Example

A beta decay followed by gamma emission

Thorium-234, written $90234Th_{90}^{234}\text{Th}$ , is neutron-rich and decays by beta-minus emission.
The mass number stays at $234$ and the atomic number rises from $90$ to $91$ , giving protactinium-234.
The equation for the first stage is $90234Th→91234Pa+−10e_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}\text{e}$ .
The protactinium nucleus is left excited, so its nucleons rearrange and it emits a gamma ray.
The equation for the second stage is $91234Pa∗→91234Pa+00γ_{91}^{234}\text{Pa}^{*} \rightarrow _{91}^{234}\text{Pa} + _{0}^{0}\gamma$ .
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.

A neutron released from a nucleus is written $01n_{0}^{1}\text{n}$ , giving it a mass number of $1$ and a charge of $0$ .
The nucleus therefore loses one nucleon, so its mass number falls by $1$ .
The neutron carries no charge, so the proton count is untouched and the atomic number stays the same.
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.
In general form, neutron emission is $ZAX→ZA−1X+01n_{Z}^{A}\text{X} \rightarrow _{Z}^{A-1}\text{X} + _{0}^{1}\text{n}$ , so oxygen-17 emitting a neutron gives $817O→816O+01n_{8}^{17}\text{O} \rightarrow _{8}^{16}\text{O} + _{0}^{1}\text{n}$ .
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 $numbersymbol_{\text{charge}}^{\text{mass number}}\text{symbol}$ 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 $ZAX_{Z}^{A}\text{X}$ .

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.

Every nucleus and particle in a nuclear equation is written as $ZAX_{Z}^{A}\text{X}$ , with the mass number $A$ on top and the charge number $Z$ below.
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.
For an emitted particle the charge number is that particle's own charge, so a beta particle takes $Z = - 1$ and a gamma ray takes $Z = 0$ .
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.
Balancing a nuclear equation therefore means finding the missing numbers that make both of those totals agree.
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.
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.

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

The method for balancing

Write the equation out with the unknown labelled, using $A$ for the missing top number and $Z$ for the missing bottom number.
Add the top numbers on the left, then add the top numbers on the right.
Set those two totals equal to each other and solve for the missing mass number.
Add the bottom numbers on the left, then add the bottom numbers on the right, remembering that a negative charge number reduces the total.
Set those two totals equal to each other and solve for the missing charge number.
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.
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 $92238U→ZATh+24α_{92}^{238}\text{U} \rightarrow _{Z}^{A}\text{Th} + _{2}^{4}\alpha$ .
Top numbers give $238 = A + 4$ .
Rearranging gives $A = 238 - 4 = 234$ .
Bottom numbers give $92 = Z + 2$ .
Rearranging gives $Z = 92 - 2 = 90$ .
The balanced equation is $92238U→90234Th+24α_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _{2}^{4}\alpha$ .
Checking gives $238 = 234 + 4$ and $92 = 90 + 2$ , so both totals agree.

Example

Finding the daughter after beta-minus emission

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

Example

Identifying an unknown emitted particle

The equation to balance is $1327Al→1227Mg+ZAX_{13}^{27}\text{Al} \rightarrow _{12}^{27}\text{Mg} + _{Z}^{A}\text{X}$ .
Top numbers give $27 = 27 + A$ , so $A = 0$ and the emitted particle contains no nucleons.
Bottom numbers give $13 = 12 + Z$ , so $Z = 1$ and the emitted particle carries one unit of positive charge.
A particle with mass number $0$ and charge $+ 1$ is a positron, written $+10e_{+1}^{0}\text{e}$ .
The balanced equation is $1327Al→1227Mg++10e_{13}^{27}\text{Al} \rightarrow _{12}^{27}\text{Mg} + _{+1}^{0}\text{e}$ .
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 $90232Th→Z1A1Ra+24α_{90}^{232}\text{Th} \rightarrow _{Z_1}^{A_1}\text{Ra} + _{2}^{4}\alpha$ followed by $Z1A1Ra→Z2A2Ac+−10e_{Z_1}^{A_1}\text{Ra} \rightarrow _{Z_2}^{A_2}\text{Ac} + _{-1}^{0}\text{e}$ .
For the first step, top numbers give $A_1 = 232 - 4 = 228$ and bottom numbers give $Z_1 = 90 - 2 = 88$ .
The first daughter is therefore $88228Ra_{88}^{228}\text{Ra}$ .
For the second step, top numbers give $A_2 = 228 - 0 = 228$ .
Bottom numbers give $Z_2 = 88 + 1 = 89$ , because the electron carries away $- 1$ .
The second daughter is $89228Ac_{89}^{228}\text{Ac}$ , so the chain runs $90232Th→88228Ra→89228Ac_{90}^{232}\text{Th} \rightarrow _{88}^{228}\text{Ra} \rightarrow _{89}^{228}\text{Ac}$ .
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 $238 = A + 4$ , 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 $1$ from the parent's charge number for a beta-minus particle, because a charge of $- 1$ on the right means the daughter must be $+ 1$ higher.
Do not give a gamma ray a mass number or a charge, since $00γ_{0}^{0}\gamma$ 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 $A$ and $Z$ in $88226Ra→ZARn+24α_{88}^{226}\text{Ra} \rightarrow _{Z}^{A}\text{Rn} + _{2}^{4}\alpha$ .
Find $A$ and $Z$ in $614C→ZAN+−10e_{6}^{14}\text{C} \rightarrow _{Z}^{A}\text{N} + _{-1}^{0}\text{e}$ .
A nucleus loses a particle of mass number $0$ and charge $- 1$ . Which particle is it?
Why does the charge number of the daughter rise when an electron is emitted?

6.3 Nuclear equations and decay

6.3.1 Beta-minus and beta-plus decay

Beta decay in unstable nuclei

Beta-minus decay

Carbon-14 decaying by beta-minus emission

Beta-plus decay

Fluorine-18 decaying by beta-plus emission

Comparing the two beta decays

Describing a beta decay for full marks

6.3.2 Effects of decay on the nucleus

The two numbers that describe a nucleus

Alpha decay

Finding the daughter of an alpha emitter

Beta-minus and beta-plus decay

Gamma emission and nuclear rearrangement

A beta decay followed by gamma emission

Neutron emission

Working out the effect of a decay

6.3.3 Balancing nuclear equations

What a nuclear equation records

The symbols you need

The method for balancing

Balancing an alpha decay

Finding the daughter after beta-minus emission

Identifying an unknown emitted particle

Balancing a decay chain of two steps

Presenting a balanced equation

Recap questions

Review 6.3 Nuclear equations and decay by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

Remember key concepts with flashcards

6.3 Nuclear equations and decay Revision Guide

6.3 Nuclear equations and decay

6.3.1 Beta-minus and beta-plus decay

Beta decay in unstable nuclei

Beta-minus decay

Carbon-14 decaying by beta-minus emission

Beta-plus decay

Fluorine-18 decaying by beta-plus emission

Comparing the two beta decays

Describing a beta decay for full marks

6.3.2 Effects of decay on the nucleus

The two numbers that describe a nucleus

Alpha decay

Finding the daughter of an alpha emitter

Beta-minus and beta-plus decay

Gamma emission and nuclear rearrangement

A beta decay followed by gamma emission

Neutron emission

Working out the effect of a decay

6.3.3 Balancing nuclear equations

What a nuclear equation records

The symbols you need

The method for balancing

Balancing an alpha decay

Finding the daughter after beta-minus emission

Identifying an unknown emitted particle

Balancing a decay chain of two steps

Presenting a balanced equation

Recap questions

Review 6.3 Nuclear equations and decay by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

Remember key concepts with flashcards

6.3 Nuclear equations and decay Revision Guide