Topic 15.8: Types of Radioactive Decay

What Topic 15.8 requires

Topic 15.8 closes Unit 15, Modern Physics and with it the AP Physics 2 course framework. The unit is weighted at 12-15% of the multiple-choice section and estimated at about 14 to 22 class periods. Where Topic 15.7 treats a sample of many nuclei statistically, 15.8 goes down to one nucleus and asks what actually happens to it.

15.8.A, describe the processes by which individual nuclei decay. One learning objective, and the essential knowledge under it is unusually specific about particles.

• 15.8.A.1 Some processes by which nuclei decay emit subatomic particles with unique properties.
• 15.8.A.1.i An alpha particle, or helium nucleus, consists of two neutrons and two protons and is symbolized by $\alpha$ or $\text{He}^{2+}$. The CED adds in brackets: "In Physics 2, only He-4 nuclei will be considered."
• 15.8.A.1.ii Neutrinos and antineutrinos have no electrical charge, have negligible mass, and are symbolized by $\nu$ and $\bar{\nu}$ respectively.
• 15.8.A.1.iii Neutrinos and antineutrinos only interact with matter via the weak force and the gravitational force, which results in very little interaction with normal matter.
• 15.8.A.1.iv Positrons, or antielectrons, have an electric charge opposite that of an electron, have the same mass as an electron, and are symbolized by $e^+$ or $\beta^+$.
• 15.8.A.2 Nuclei can undergo radioactive decay via alpha decay, beta-minus decay, beta-plus decay and gamma decay.
• 15.8.A.2.i In all nuclear decays, nucleon number (the number of neutrons and protons), lepton number (the number of electrons and neutrinos), and charge are conserved.
• 15.8.A.2.ii Alpha decay occurs when a nucleus ejects an alpha particle.
• 15.8.A.2.iii Beta-minus decay occurs when a neutron changes to a proton by emitting an electron and antineutrino.
• 15.8.A.2.iv Beta-plus decay occurs when a proton changes to a neutron by emitting a positron and neutrino.
• 15.8.A.2.v Gamma decay occurs after a nucleus has undergone alpha or beta decay and the excited nucleus decays to a lower energy state by emitting a photon.
• 15.8.A.3 The type of decay exhibited by a given nucleus is determined by the isotope of the element.

The boundary statement, in full, because every clause of it changes what you should study:

"AP Physics 2 does not expect students to memorize the processes by which specific isotopes decay or the half-lives of specific isotopes. Neutron emission and electron capture are not included in the AP Physics 2 curriculum framework. Additionally, types of neutrinos, the characteristics that distinguish neutrinos and antineutrinos, and an explanation or application of the weak force are not within the scope of this course."

Three separate exclusions there, and they are the reason this topic is a method rather than a list.

• No memorising which isotope does what. Combined with 15.8.A.3, which says the isotope determines the decay type, this tells you the exam will hand you the decay type or enough information to deduce it, and never expect recall.
• Neutron emission and electron capture are out. Four decay modes, not six.
• The weak force is named but not explained. 15.8.A.1.iii tells you neutrinos interact via the weak force, and the boundary statement then puts any explanation or application of that force out of scope. Know the name, do not build an argument on the mechanism.

Four suggested skills: 1.A create diagrams, tables, charts, or schematics to represent physical situations; 2.C compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Two conservation laws do all the work

There is a version of this topic that is four facts to be memorised and forgotten. There is a better version in which there is nothing to memorise at all, and 15.8.A.2.i is what makes it possible: in all nuclear decays, nucleon number, lepton number and charge are conserved.

Write a decay as an equation and every product is forced. Nuclear notation, which the CED names in 15.2.A.1.iii without printing a format, writes a nuclide with its mass number above and its atomic number below:

• $A$ is the mass number or nucleon number, the total count of protons and neutrons.
• $Z$ is the atomic number, the number of protons, which for a bare nucleus is also its charge in units of the elementary charge.
• The neutron count is $A - Z$, which is worth writing down whenever a question asks about neutrons rather than nucleons.

Now assign $A$ and $Z$ to every particle a decay can emit, and the bookkeeping closes.

Every entry in that table comes straight from the CED. The alpha is two protons and two neutrons, so four nucleons and charge $+2$ (15.8.A.1.i). The positron has charge opposite an electron's, so $+1$ (15.8.A.1.iv). Neutrinos have no electrical charge (15.8.A.1.ii). Nothing here needs remembering; it needs reading.

Then two sums balance, every time.

1. Mass numbers: the parent's $A$ equals the sum of the products' $A$ values.
2. Charges: the parent's $Z$ equals the sum of the products' $Z$ values.

That is the whole method. Given any three of the four things in a decay, the fourth follows from those two equations, and you never have to recall which decay does what to $Z$. The table in the next section is a summary of results you can rebuild in ten seconds, not a list to memorise.

A note on the third conservation law. 15.8.A.2.i also lists lepton number, and its parenthetical gloss is "the number of electrons and neutrinos". Taken literally that count is not conserved in beta-minus decay, which starts with neither and produces two particles. It balances once antiparticles are counted with the opposite sign: an electron and a neutrino each count $+1$, a positron and an antineutrino each count $-1$. That is exactly why the CED pairs an electron with an antineutrino in 15.8.A.2.iii and a positron with a neutrino in 15.8.A.2.iv rather than the other way round. The same boundary statement then puts "the characteristics that distinguish neutrinos and antineutrinos" out of scope, so treat the signed lepton count as the reason the pairings are what they are, and use nucleon number and charge as the checks you actually run. Charge conservation is the same law as Topic 10.2, applied to a nucleus.

The four decays in one table

Here is everything 15.8.A.2 requires, in the form a question will ask for it. Each row is derivable from the previous section; none of it needs to be held in memory.

Four readings of that table are worth having.

Only alpha decay changes $A$. Both betas move a nucleon between the two types without changing the total, and gamma moves no nucleons at all. So if a question tells you the mass number changed, alpha decay was involved and you can say how many alphas: the drop in $A$ divided by 4.

The two betas are mirror images. Beta-minus raises $Z$ by one, beta-plus lowers it by one. The rest of the mirroring follows: neutron to proton against proton to neutron, electron against positron, antineutrino against neutrino.

Beta-minus raising the atomic number is the row students reverse. The name has a minus in it and the atomic number goes up. The minus refers to the charge of the emitted particle, not to the change in $Z$. Charge conservation is the fix: the nucleus threw out a negative charge, so what remains must be more positive than before.

Gamma decay changes neither number, which is why 15.7.B.1 was careful to define radioactive decay as transformation into different nuclei or to a lower energy level of the same nucleus. Gamma is the second clause of that definition.

Two general observations to carry into any question about a decay chain. Alpha decay reduces $Z$ twice as fast as it reduces the neutron count, so it leaves the remaining nucleus relatively neutron-rich. Beta-minus then pushes back the other way, converting a neutron into a proton. That interplay is why real decay series alternate between the two, and why a chain of one alpha and two beta-minus decays returns to the original element with four fewer nucleons, which is the third worked example below.

Alpha decay

Essential knowledge 15.8.A.2.ii is one sentence: alpha decay occurs when a nucleus ejects an alpha particle. Everything else is in 15.8.A.1.i, which describes the particle itself.

The particle. Two neutrons and two protons, which is a helium nucleus. The CED symbolises it as $\alpha$ or $\text{He}^{2+}$, and note what that second symbol says: charge $+2$, because there are two protons and no electrons attached. The bracketed restriction is worth reading twice: "In Physics 2, only He-4 nuclei will be considered." So the alpha particle in every AP Physics 2 question has $A = 4$ and $Z = 2$, with no exceptions to watch for.

The effect on the parent. Take away four nucleons of which two are protons:

Both sums balance by construction. Mass numbers: $A = (A - 4) + 4$. Charges: $Z = (Z - 2) + 2$. That is the entire derivation, and it is faster than recalling the row from a table.

What the daughter is called. A question can ask you for the daughter's mass number and atomic number and you can always answer. It can ask for the element's name only if it gives you a way to look up the atomic number, because the CED's appendix contains no periodic table. The AP Physics 2 Table of Information prints constants and conversion factors, unit symbols, prefixes, trigonometric values for common angles, the exam conventions, the equation tables and a geometry and trigonometry table, and that is all of it. Answer in $A$ and $Z$ unless a question hands you the elements.

A physical detail worth one line. The alpha carries away $+2e$ of charge and four nucleons, so momentum conservation sends the much heavier daughter nucleus recoiling in the opposite direction, slowly. That is Topic 4.3 applied to a nucleus, and 15.7.A.3 lists conservation of momentum as one of the constraints on nuclear processes, so it is in scope.

Beta-minus and beta-plus, with the signs exactly right

The two beta decays are where sign errors live, so here is each one written out and checked against both conservation laws.

Beta-minus decay. Essential knowledge 15.8.A.2.iii: a neutron changes to a proton by emitting an electron and antineutrino. At the level of the single nucleon,

and at the level of the nucleus,

• Mass numbers: $A = A + 0 + 0$. Nothing left the nucleus that counted as a nucleon; a neutron simply became a proton.
• Charges: $Z = (Z + 1) + (-1) + 0$. This is the check that gets the direction right. Throwing out a negative particle leaves the remainder more positive, so $Z$ goes up.
• Result: the atomic number rises by one, the mass number holds, and the neutron count falls by one.

Beta-plus decay. Essential knowledge 15.8.A.2.iv: a proton changes to a neutron by emitting a positron and neutrino. So

and

• Mass numbers: $A = A + 0 + 0$, for the same reason as before.
• Charges: $Z = (Z - 1) + (+1) + 0$. A positive particle left, so the remainder is less positive and $Z$ goes down.
• Result: the atomic number falls by one, the mass number holds, and the neutron count rises by one.

The pairing of the light particles is not arbitrary and not interchangeable. Beta-minus emits an electron with an antineutrino; beta-plus emits a positron with a neutrino. The CED states both pairings explicitly, in 15.8.A.2.iii and 15.8.A.2.iv, and swapping them is a factual error even though neither particle affects $A$ or $Z$. The reason is the lepton-number clause of 15.8.A.2.i, discussed above: each emitted lepton has to be balanced by an antilepton for the count to hold at zero.

Why neutrinos are in the course at all. They carry no charge and negligible mass (15.8.A.1.ii), so they change nothing in the $A$ and $Z$ bookkeeping and could be dropped from a balanced equation without breaking it. What they do carry is energy and momentum, and 15.7.A.3 makes both conservation laws constraints on nuclear processes. 15.8.A.1.iii then explains why they are so hard to detect: they interact only via the weak force and gravity, so almost nothing stops them. Note the fence around that statement, though. The boundary statement puts any explanation or application of the weak force outside the course, so the required knowledge is "they barely interact", not a mechanism.

Gamma decay, and when it happens

Essential knowledge 15.8.A.2.v is more specific than students usually notice. Read it slowly: gamma decay occurs after a nucleus has undergone alpha or beta decay, and the excited nucleus decays to a lower energy state by emitting a photon.

Two things are being said.

Gamma decay is a follow-on, not a starting move. In the CED's framing it is what an already-decayed nucleus does to shed leftover energy, so a gamma appears in a chain after an alpha or a beta, and it does not initiate one.

Nothing about the nucleus's identity changes. No nucleons leave, no charge leaves. The photon carries away energy and nothing else that the bookkeeping tracks:

with the asterisk marking the excited state. Both sums balance trivially, since a photon has $A = 0$ and $Z = 0$.

This is the second half of the definition in 15.7.B.1, which called radioactive decay the spontaneous transformation of a nucleus into one or more different nuclei or to a lower energy level of the same nucleus. That "or" clause exists for gamma decay.

The parallel with atomic physics is exact and worth using, because it makes the energetics obvious. In Topic 15.3, an atom in an excited state drops to a lower level and emits a photon whose energy equals the gap between the levels. A nucleus does the same thing with nuclear energy levels instead of electron levels. The photon energy still follows the sheet's $E = hf$, and the gaps are simply enormous by comparison, which is why the photons come out as gamma rays rather than as visible light.

One practical consequence for a 2.C comparison question. If you are told a nucleus emitted a gamma ray and asked how its mass number and atomic number changed, the answer is that neither changed. If you are told the same nucleus also emitted an alpha at some point in the process, the alpha did all the changing and the gamma did none of it.

Balancing any decay equation: the method

Skill 1.A asks for representations and skills 3.B and 3.C ask for claims backed by a law. In this topic the law is 15.8.A.2.i and the representation is a balanced equation, so one routine covers almost every question.

1. Write both sides with $A$ and $Z$ on every particle, including the light ones, using the particle table above. A neutrino written with no numbers on it is fine, but know that they are zeros.
2. Balance mass numbers. The top numbers must sum to the same value on both sides.
3. Balance charges. The bottom numbers must sum to the same value on both sides.
4. Name the decay from what the change in $Z$ and $A$ turned out to be, using the four-row table, rather than assuming a type at the start.
5. State the law you used. A 3.B answer that says "conservation of nucleon number and conservation of charge" scores; one that says "because that is what alpha decay does" is asserting the thing being asked.

That routine also runs backwards, which is how most multiple-choice questions on this topic are built. Given a parent and a daughter, subtract:

• $A$ fell by 4 and $Z$ fell by 2, so it was alpha.
• $A$ unchanged and $Z$ rose by 1, so it was beta-minus.
• $A$ unchanged and $Z$ fell by 1, so it was beta-plus.
• Neither changed, so it was gamma.

Six errors worth naming directly.

• Beta-minus lowering the atomic number. It raises it. The minus is the charge of the emitted electron, not the change in $Z$.
• Alpha decay changing $A$ by 2. It changes $Z$ by 2 and $A$ by 4. Two protons and two neutrons left, so four nucleons left.
• Gamma decay changing the element. It changes neither number, only energy.
• Pairing an electron with a neutrino. Beta-minus emits an electron with an antineutrino; beta-plus emits a positron with a neutrino. Both pairings are stated in the CED.
• Reaching for electron capture or neutron emission. The boundary statement excludes both from AP Physics 2.
• Trying to recall which isotope decays how. The boundary statement says you are not expected to, and 15.8.A.3 says only that the isotope determines the type, not which type.

On exam shape: the AP Physics 2 exam runs 3 hours with 42 multiple-choice questions and 4 free-response questions, weighted 50/50, and a calculator is allowed on both sections. This topic's four suggested skills are all representation and argument skills, with no calculation skill among them, which is a clear signal about the form its questions take. Expect to balance an equation, compare two decays, or justify a claim, rather than to compute a number.

For the sample as a whole, half-life and the energy released, see Topic 15.7. For nuclear structure and nuclear notation, see Topic 15.2. For charge conservation as a general law, see Topic 10.2. The Unit 15 overview sets the eight topics in order and the AP Physics 2 equation sheet holds the ten Modern Physics entries, none of which is needed to balance a decay equation.