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Balancing a decay the way you balanced a reaction, with the mass number and the atomic number conserved instead of the atoms of each element.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write a nuclear equation for alpha decay, beta-minus decay and gamma emission, and to work out the daughter nuclide from the parent and the emitted particle rather than remembering it. You will be able to audit any nuclear equation by totaling the mass number and the atomic number on each side, work backwards from a daughter to the parent that produced it, and say why the element changes here when a chemical equation may never change it.
You can read the mass number and the atomic number off a nuclear symbol, and you can predict from those two numbers whether a nuclide will shed an alpha particle or turn a neutron into a proton. You also balanced chemical equations earlier in this course, and you know why only the numbers in front may change. This lesson writes a decay down, and the writing turns out to be the same skill with a different quantity being counted.
| Term | What it means |
|---|---|
| Nuclear equation | The decaying nucleus on the left; what is left and what was emitted on the right. |
| Parent | The nucleus that decays. |
| Daughter | The nucleus left behind. |
| Alpha particle | $\mathrm{^{4}_{2}He}$: two protons and two neutrons. |
| Beta-minus particle | $\mathrm{^{0}_{-1}e}$: an electron from the nucleus. |
| Gamma ray | $\mathrm{^{0}_{0}\gamma}$: energy with no particles. |
The second lesson of this course said that a chemical equation must balance because atoms are rearranged and never created, and that the counts to compare are the atoms of each element on each side.
A nuclear equation breaks that flatly. Uranium goes in and thorium comes out. The uranium atom has not been rearranged; it has stopped being uranium.
Conservation has not stopped applying. What is conserved has changed, and there are now two things to count rather than one per element:
Both are conserved exactly. And that is enough to work out the daughter without being told it, which is the point of the whole lesson: you are never asked to remember what uranium-238 turns into, you are asked to subtract.
The three emitted particles are written with their own two numbers so that they can go into the sum:
| Particle | Written | Mass number | Atomic number |
|---|---|---|---|
| alpha | $\mathrm{^{4}_{2}He}$ | 4 | 2 |
| beta-minus | $\mathrm{^{0}_{-1}e}$ | 0 | −1 |
| gamma | $\mathrm{^{0}_{0}\gamma}$ | 0 | 0 |
The beta-minus particle's $-1$ is the one that surprises people. An electron has one negative charge, so its atomic number is $-1$; subtracting $-1$ from the parent's atomic number adds one, which is exactly what happens when a neutron inside the nucleus becomes a proton.
Another way: picture
Think of two cash registers, one counting particles and one counting charge, and a nucleus walking past both on its way out of the store. Whatever it leaves behind and whatever it takes with it must ring up the same totals as it had going in. The alpha particle rings up 4 and 2; the beta-minus particle rings up 0 and minus 1; the gamma ray rings up nothing at all.
Another way: steps
To write a nuclear equation:
Write the parent in full. Mass number on top, atomic number below. If only a name is given, take the mass number from the name and the atomic number from the periodic table.
Write the emitted particle in full. Alpha $^{4}_{2}$, beta-minus $^{0}_{-1}$, gamma $^{0}_{0}$. Writing the little numbers is not optional; they are what goes into the arithmetic.
Subtract for the daughter. Parent's mass number minus particle's; parent's atomic number minus particle's.
Name the daughter from its atomic number.
For a chain, repeat — one decay at a time, writing each middle nuclide down.
To work backwards, add instead: daughter plus particle gives the parent.
Check the work. Add the right-hand side. Both totals must equal the parent's two numbers exactly. Then a sense check: after alpha decay the element should be two places to the left in the periodic table; after beta-minus, one place to the right; after gamma, unchanged. And no daughter should have a negative neutron count or an atomic number bigger than its mass number.
Conserving the mass number is allowed because protons and neutrons are neither created nor destroyed in these decays; at most a neutron becomes a proton, and both count toward the mass number.
Conserving the atomic number is allowed because it is really a count of charge, and charge is conserved in every process known. When a neutron becomes a proton, the electron that leaves carries away exactly the charge the new proton added.
Giving the beta-minus particle an atomic number of $-1$ is required for that bookkeeping: its charge is one negative unit, so in a sum of charges it counts as $-1$.
Subtracting to find the daughter is allowed because, with the totals fixed, the daughter is whatever remains once the particle's share is taken out.
Naming the daughter from its atomic number alone is allowed because the atomic number is the definition of an element. Two nuclei with the same atomic number are the same element, whatever their mass numbers.
Here is the audit of a chemical equation, $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$:
| Counted | Left | Right |
|---|---|---|
| $\mathrm{H}$ atoms | 4 | 4 |
| $\mathrm{O}$ atoms | 2 | 2 |
And here is the audit of a nuclear one, $\mathrm{^{238}_{92}U} \rightarrow \mathrm{^{234}_{90}Th} + \mathrm{^{4}_{2}He}$:
| Counted | Left | Right |
|---|---|---|
| mass number | 238 | $234 + 4 = 238$ |
| atomic number | 92 | $90 + 2 = 92$ |
The same table, the same discipline, a different pair of rows. And one difference that matters: in the first table the row labels name elements, and in the second they do not — which is precisely why the element is free to change in the second and not in the first.
It is worth being blunt about the thing that feels wrong here.
Everything earlier in this course insisted that a reaction may not turn one element into another. Lead does not become gold; that is why the alchemists failed, and it is why changing a subscript to force an equation to balance is forbidden. All of that is still true of chemistry, because a chemical change only ever moves electrons around and leaves every nucleus exactly as it found it.
A nuclear change reaches the nucleus, and the nucleus is where the element lives. So lead really can become gold, and uranium really does become thorium. What a nuclear change cannot do is lose a proton or a neutron out of the accounts altogether — and that is the conservation this lesson grades.
The honest summary is that the two rules were never in competition. Atoms of each element are conserved is a statement about chemistry. Mass number and atomic number are conserved is the more general statement, and chemistry is the special case of it where the nuclei never change, so counting elements works.
Many American schools, theaters and airplanes have EXIT signs that glow without any wiring at all, and the light comes from a nuclear equation. The sign holds sealed glass tubes of tritium, hydrogen-3, coated inside with a phosphor.
Tritium has one proton and two neutrons — too many neutrons for so light a nucleus — so it decays by beta-minus emission: $\mathrm{^{3}_{1}H} \rightarrow \mathrm{^{3}_{2}He} + \mathrm{^{0}_{-1}e}$. The mass number stays at 3; the atomic number goes from 1 to 2, so hydrogen becomes helium. The audit checks: $3 + 0 = 3$ and $2 + (-1) = 1$. The emitted electrons strike the phosphor and make it glow, around the clock, for years, through any power failure — exactly when an exit sign is needed most.
The beta particles are weak enough that the glass tube stops them completely, so the sign is safe to stand under. But the Nuclear Regulatory Commission still licenses these signs and requires that they be returned for disposal rather than thrown away, because a broken tube releases tritium gas. Tritium's half-life is about 12 years, so a sign's glow fades noticeably over a decade or two — the subject of the next lesson.
Every living thing takes in carbon-14 from the air. After death, $\mathrm{^{14}_{6}C} \rightarrow \mathrm{^{14}_{7}N} + \mathrm{^{0}_{-1}e}$ steadily turns it into nitrogen. Measuring how much carbon-14 is left dates bones, wood and cloth up to about 50,000 years old.
Treating a nuclear change like a chemical one. A nuclear change conserves two counts rather than one: the mass number, which is protons plus neutrons, and the atomic number, which is protons alone. Balancing a nuclear equation is the same move as balancing a chemical one with a different quantity conserved — and because the atomic number changes, the element changes, which is the one thing a chemical equation may never do.
The emitted particle is left out of the sum. The commonest error by a distance: a learner writes the daughter as having the parent's numbers, because the particle that left seemed too small to matter. An alpha particle takes four units of mass number away, which is a great deal; a beta-minus particle takes none, which is why it is easy to think it takes none of anything, and its atomic number of $-1$ still moves the element along by one.
The sign on the beta-minus particle is read as a mistake. An atomic number of $-1$ looks wrong on a particle. It is not: the atomic number counts charge in units of a proton's charge, and an electron has one negative charge. Subtract it and the daughter's atomic number goes up, which is the correct and slightly counterintuitive answer.
A chain is done in one step. Uranium-238 reaches lead-206 after fourteen decays, not one, and the intermediate nuclides are all unstable too. When two decays are given, do them one at a time and write the middle nuclide down. Applying two changes to one starting pair of numbers at once is how a learner ends up applying one of them twice.
The equation is thought to need a coefficient. It does not. Every coefficient in a nuclear equation of this kind is one, and the only thing worth working out is which nuclide the daughter is.
Write the parent in full.
$\mathrm{^{226}_{88}Ra}$
Both numbers before any arithmetic.
Write the alpha particle.
$\mathrm{^{4}_{2}He}$
Its own two numbers.
Subtract the mass numbers.
$226 - 4 = 222$
What is left.
Subtract the atomic numbers.
$88 - 2 = 86: \text{radon}$
The periodic table names it.
Write and check the equation.
$\mathrm{^{226}_{88}Ra} \rightarrow \mathrm{^{222}_{86}Rn} + \mathrm{^{4}_{2}He}$
$222 + 4 = 226$ and $86 + 2 = 88$.
Write the parent in full.
$\mathrm{^{14}_{6}C}$
Mass number 14, atomic number 6.
Write the beta-minus particle.
$\mathrm{^{0}_{-1}e}$
Its numbers are 0 and minus 1.
Subtract the mass numbers.
$14 - 0 = 14$
Unchanged.
Subtract the atomic numbers.
$6 - (-1) = 7$
Subtracting a negative adds.
Name the daughter.
$Z = 7: \text{nitrogen}$
Same size, different element.
Check the totals.
$14 + 0 = 14; \ 7 + (-1) = 6$
The audit run backwards.
Write the parent in full.
$\mathrm{^{222}_{86}Rn}$
Radon from a basement.
Apply the first decay.
$\text{alpha: } 222 - 4, \ 86 - 2$
Too heavy, so alpha.
Write the middle nuclide.
$\mathrm{^{218}_{84}Po}$
Polonium-218, written down.
Apply the second decay.
$\text{alpha again: } 218 - 4, \ 84 - 2$
Still above 82.
Write the end nuclide.
$\mathrm{^{214}_{82}Pb}$
Lead-214.
Check the whole chain.
$214 + 4 + 4 = 222; \ 82 + 2 + 2 = 86$
Both totals conserved.
Note it is still unstable.
$\text{lead-214 is neutron-rich}$
The chain continues by beta decay.
Subtract the mass numbers.
$234 - 0 = 234$
Nothing with a mass number has left.
Subtract the atomic numbers.
$90 - (-1) = 91: \text{protactinium}$
Subtracting a negative adds.
Write the equation.
A nucleus of potassium-40 undergoes beta-minus decay and what is left is calcium-40. Which of these describes what has happened?
Complete the worked solution: polonium-210, used in the static eliminators on photographic and industrial equipment, has mass number two hundred ten and atomic number eighty-four, and it sheds an alpha particle. Find the daughter's mass number, its atomic number, and its neutrons.
Lower the mass number.
$\text{two hundred ten} - \text{four} =$ a
The alpha particle's mass number.
Lower the atomic number.
$\text{eighty-four} - \text{two} =$ z
Lead, the end of the line.
Count the daughter's neutrons.
$(\text{mass number}) - (\text{atomic number}) =$ n
A stable lead nucleus, the end of the chain.
polonium-210 undergoes alpha decay. Conservation decides the daughter's two numbers — you do not get to choose them. Give them.
$$\mathrm{^{210}_{84}Po}$ \ \rightarrow \ \mathrm{^{A}_{Z}X} \ + \ $\mathrm{^{4}_{2}He}$$, so $A =$ a and $Z =$ b.
A nucleus somewhere in a rock has decayed. All anyone found afterwards was an alpha particle and a nucleus of lead-214. Work backwards: what decayed, and how many neutrons did it have?
| mass number | atomic number | neutrons | |
|---|---|---|---|
| the nucleus that decayed | |||
| $\mathrm{^{4}_{2}He}$ | 4 | 2 | — |
| $\mathrm{^{214}_{82}Pb}$ | 214 | 82 |
strontium-90, $\mathrm{^{90}_{38}Sr}$, is found in the fallout from 1950s weapons tests, still traced in soil. It decays by beta-minus emission. How many neutrons does the daughter nucleus have?
The answer: a.
One decay almost never reaches a stable nucleus. In a lump of uranium ore in a university geology store, a nucleus of uranium-238 undergoes alpha decay, and what it becomes then undergoes beta-minus decay. Give the two numbers of the nucleus left at the end.
After both decays the nucleus left has mass number a and atomic number b.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
carbon-14 undergoes beta-minus decay, giving nitrogen-14 and a beta-minus particle. Audit the equation: fill in the two numbers for the particle that leaves and for the nucleus that is left, then total the right-hand side. If the equation is a true statement, the totals must match the parent exactly.
| mass number | atomic number | |
|---|---|---|
| $\mathrm{^{14}_{6}C}$ | 14 | 6 |
| $\mathrm{^{0}_{-1}e}$ | ||
| $\mathrm{^{14}_{7}N}$ | ||
| the two totals on the right |
You can find a daughter nuclide by subtracting the emitted particle's two numbers, and you can check an equation by totaling both counts on each side. Tell someone why an electron leaving a nucleus makes its atomic number go up rather than down. Next: how long all of this takes, and why nothing you can do to a sample changes the answer.
15. Your turn: thorium-234, $\mathrm{^{234}_{90}Th}$, undergoes beta-minus decay. Write the equation., step 3
$\mathrm{^{234}_{90}Th} \rightarrow \mathrm{^{234}_{91}Pa} + \mathrm{^{0}_{-1}e}$
Check: $234 + 0$ and $91 + (-1)$.