Nuclear Decay

AP Physics 2· AP Physics 2 CED — Quantum, Atomic, and Nuclear Physics· 14 min read

1. Nuclear Decay Fundamentals and Notation★★☆☆☆⏱ 3 min

Nuclear decay (also called radioactive decay) is the spontaneous breakdown of an unstable atomic nucleus, which emits ionizing radiation to reach a more stable lower-energy state. Nuclei become unstable due to an incorrect neutron-to-proton ratio, excessive total nucleon count, or after being left in an excited state from a prior nuclear reaction. While it is impossible to predict when any single nucleus will decay, large collections of unstable nuclei follow predictable statistical behavior.

📘 Definition

Nuclide Notation

^A_Z \text{X}

Standard notation for representing an atomic nucleus, where $A$ is the mass (nucleon) number (total protons + neutrons), $Z$ is the atomic (proton/charge) number, and $X$ is the element's chemical symbol.

2. Types of Decay and Conservation Laws★★☆☆☆⏱ 4 min

All nuclear decay reactions must satisfy two fundamental conservation laws tested consistently on the AP Physics 2 exam: conservation of total nucleon number ( $A$ ) and conservation of total electric charge ( $Z$ ). The sum of $A$ values on the reactant side equals the sum of $A$ values on the product side, and the same equality holds for $Z$ values.

Alpha decay: Occurs for very heavy nuclides ( $A > 200$ ) that are too large to be stable. Emits an alpha particle ( $_{2}^{4} α$ , identical to a helium nucleus), resulting in transmutation to a new element.
Beta-minus decay: The most common beta decay type, occurring when a nucleus has too many neutrons. A neutron decays into a proton, a high-energy electron ( $_{- 1}^{0} β^{-}$ ), and an antineutrino, resulting in transmutation.
Gamma decay: Occurs when a nucleus is in an excited energy state. Emits a high-energy gamma photon ( $_{0}^{0} γ$ ) to release excess energy, with no change to $A$ or $Z$ , so no transmutation occurs.

📐 Worked Example

Carbon-14 ( $_{6}^{14} C$ ) undergoes beta-minus decay to form a nitrogen (N) daughter nuclide. Write the complete balanced decay equation and identify the mass and atomic numbers of nitrogen.

1
Start with the general beta-minus decay skeleton:
2
$^{14}_{6} \text{C} \rightarrow ^A_{Z} \text{N} + ^0_{-1} \beta^- + \bar{\nu}$
3
Apply conservation of nucleon number to solve for $A$ :
4
$14 = A + 0 \implies A = 14$
5
Apply conservation of charge number to solve for $Z$ :
6
$6 = Z + (-1) \implies Z = 7$
7
Write the full balanced equation, confirming the daughter nuclide is nitrogen-14:
8
$^{14}_{6} \text{C} \rightarrow ^{14}_{7} \text{N} + ^0_{-1} \beta^- + \bar{\nu}$

3. Exponential Decay, Half-Life, and Activity★★★☆☆⏱ 5 min

Radioactive decay is a statistical process where the instantaneous rate of decay is proportional to the number of undecayed nuclei remaining $N (t)$ . This gives the differential decay law $\frac{d N}{d t} = - λ N$ , where $λ$ is the decay constant (units of inverse time, larger $λ$ means faster decay). Solving this gives the exponential decay law:

N(t) = N_0 e^{-\lambda t}

Where $N_{0}$ is the initial number of undecayed nuclei at $t = 0$ . Half-life ( $T_{1/2}$ ) is defined as the time required for half of the original unstable nuclei to decay. Substituting $N = N_{0} /2$ at $t = T_{1/2}$ gives the core relation between half-life and decay constant:

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}

Activity $R (t)$ is the measurable decay rate (number of decays per unit time), equal to $R = \frac{d N}{d t} = λ N (t)$ . Activity also follows the exponential decay law $R (t) = R_{0} e^{- λ t}$ , where $R_{0} = λ N_{0}$ is initial activity. For calculations, it is often easier to write decay directly in terms of half-life: after $n = t / T_{1/2}$ half-lives, $N (t) = N_{0} (\frac{1}{2})^{t / T_{1/2}}$ and $R (t) = R_{0} (\frac{1}{2})^{t / T_{1/2}}$ .

📐 Worked Example

A sample of radioactive iodine-131 has an initial activity of 1280 MBq. Iodine-131 has a half-life of 8 days. What is the activity of the sample after 24 days? Calculate the initial number of undecayed iodine-131 nuclei, given $λ = 0.0866 day^{- 1}$ .

1
Calculate the number of half-lives passed:
2
$n = \frac{t}{T_{1/2}} = \frac{24}{8} = 3$
3
Use the half-life activity formula to find final activity:
4
$R = R_0 \left(\frac{1}{2}\right)^n = 1280 \times \left(\frac{1}{2}\right)^3 = 160 \text{ MBq}$
5
To find initial number of nuclei, use $R_{0} = λ N_{0}$ , convert activity to decays per day to match units of $λ$ :
6
$1280 \text{ MBq} = 1.10592 \times 10^{14} \text{ decays/day}$
7
Solve for $N_{0}$ :
8
$N_0 = \frac{R_0}{\lambda} = \frac{1.10592 \times 10^{14}}{0.0866} \approx 1.28 \times 10^{15} \text{ nuclei}$

4. Mass-Energy Equivalence and Decay Q-Value★★★☆☆⏱ 4 min

For a nuclear decay to be spontaneous, the total mass of the decay products must be less than the mass of the parent nuclide. The mass difference $Δ m$ is converted to kinetic energy of the decay products, per $E = Δ m c^{2}$ . This energy is called the Q-value of the decay, defined as:

Q = (m_{\text{parent}} - \sum m_{\text{products}}) c^2

If $Q > 0$ , decay is spontaneous (exothermic), which is true for all naturally occurring nuclear decay. If $Q < 0$ , decay cannot occur spontaneously. A convenient shortcut for AP problems: use atomic masses (not nuclear masses), because the total number of electrons is the same on both sides of alpha and beta-minus decay, so electron masses cancel out. Recall that $1 u \cdot c^{2} = 931.5 MeV$ .

📐 Worked Example

Polonium-210 ( $_{84}^{210} Po$ , atomic mass = 209.98287 u) undergoes alpha decay to form lead-206 ( $_{82}^{206} Pb$ , atomic mass = 205.97447 u) and an alpha particle ( $_{2}^{4} He$ , atomic mass = 4.00260 u). Calculate the Q-value of the decay and confirm it is spontaneous.

1
Confirm the decay equation is balanced, and note that electron masses cancel:
2
$^{210}_{84} \text{Po} \rightarrow ^{206}_{82} \text{Pb} + ^4_2 \text{He} \quad (84 e^- = 82 e^- + 2 e^-)$
3
Calculate total mass of the products:
4
$m_{\text{products}} = 205.97447 + 4.00260 = 209.97707 \text{ u}$
5
Calculate the mass difference between parent and products:
6
$\Delta m = m_{\text{parent}} - m_{\text{products}} = 209.98287 - 209.97707 = 0.00580 \text{ u}$
7
Convert mass difference to Q-value using the 1 u = 931.5 MeV/c² conversion:
8
$Q = 0.00580 \text{ u} \cdot c^2 \times 931.5 \frac{\text{MeV}}{\text{u} \cdot c^2} \approx 5.40 \text{ MeV}$
9
Since $Q > 0$ , the decay is spontaneous, as expected.

5. AP Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding of nuclear decay with these AP-style questions:

A researcher has a pure sample of radioactive sodium-24, which has a half-life of 15 hours. The initial activity of the sample is 800 Bq. What is the activity of the sample after 60 hours?
- A) 100 Bq
- B) 200 Bq
- C) 50 Bq
- D) 400 Bq
Reveal answer
C) 50 Bq —
Correct: 4 half-lives have passed, so $R = 800 \times (1/2)^{4} = 50$ Bq. All other options correspond to incorrect counts of half-lives.
A geologist is dating a volcanic rock sample that contains potassium-40, which has a half-life of $1.25 \times 1 0^{9}$ years. The ratio of undecayed potassium-40 to decay product argon-40 in the sample is 1:3. What is the age of the rock?
- A) $1.25 \times 1 0^{9}$ years
- B) $2.5 \times 1 0^{9}$ years
- C) $3.75 \times 1 0^{9}$ years
- D) $5.0 \times 1 0^{9}$ years
Reveal answer
B) $2.5 \times 10^9$ years —
1/4 of the original potassium remains, which equals 2 half-lives, so age = $2 \times 1.25 \times 1 0^{9} = 2.5 \times 1 0^{9}$ years.
A hospital orders a 500 MBq sample of technetium-99m for a cardiac imaging scan. Technetium-99m has a half-life of 6 hours. If the sample is prepared 24 hours before it is used, what is the activity when it is used, and does it meet the minimum 25 MBq requirement?
Reveal answer
31.25 MBq, meets the requirement —
4 half-lives passed, so $R = 500/16 = 31.25$ MBq, which is above the 25 MBq minimum, so the sample is usable.

6. Common Pitfalls

Wrong move:

Balancing beta-minus decay by decreasing the daughter mass number $A$ by 1 and atomic number $Z$ by 1 to account for the emitted electron.

Why:

Students confuse the tiny mass of an electron with a nucleon, and mix up the sign of the beta particle’s charge.

Correct move:

Remember beta-minus decay converts a neutron to a proton, so total $A$ stays the same, and $Z$ increases by 1.

Wrong move:

Using the inverse relation $λ = \frac{T _{1/2}}{l n 2}$ instead of $T_{1/2} = \frac{l n 2}{λ}$ , leading to a factor of ~2 error in calculations.

Why:

Students misremember which quantity goes in the numerator of the half-life/decay constant relation.

Correct move:

Always verify your result with the $(1/2)^{n}$ half-life rule to cross-check for algebra errors.

Wrong move:

Calculating Q-value as $Q = (\sum m_{products} - m_{parent}) c^{2}$ , getting a negative Q even for spontaneous decay.

Why:

Students mix up the definition of Q-value with mass defect for binding energy.

Correct move:

Always write Q as (initial mass minus final mass) for decay, since the missing mass is converted to kinetic energy of products.

Wrong move:

Claiming that after two half-lives, all original unstable nuclei have decayed.

Why:

Students misinterpret half-life as the total lifetime of the entire sample.

Correct move:

Remember half-life is the time for half of the remaining nuclei to decay, so 1/4 of the original sample remains after 2 half-lives, 1/8 after 3, etc.

Wrong move:

Changing the $A$ or $Z$ number of the parent nuclide for gamma decay.

Why:

Students assume all decay changes the nuclide identity, forgetting gamma is only energy emission.

Correct move:

Always write gamma as $_{0}^{0} γ$ , so $A$ and $Z$ of the daughter are identical to the parent.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Nuclide Notation	$_{Z}^{A} X$	$A$ = total nucleon (mass) number, $Z$ = proton (charge) number
Conservation Laws	$\sum A_{left} = \sum A_{right}$ , $\sum Z_{left} = \sum Z_{right}$	Applies to all nuclear decay reactions
Alpha Decay General Form	$_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} α$	Occurs in heavy unstable nuclides
Beta-Minus Decay General Form	$_{Z}^{A} X \to_{Z + 1}^{A} Y +_{- 1}^{0} β^{-} + \overset{ν}{ˉ}$	Occurs in nuclides with too many neutrons
Exponential Decay Law	$N (t) = N_{0} e^{- λ t}$ , $R (t) = R_{0} e^{- λ t}$	$N$ = undecayed nuclei, $R$ = activity, $λ$ = decay constant
Half-Life / Decay Constant	$T_{1/2} = \frac{l n 2}{λ} = \frac{0.693}{λ}$	Relates the two common decay parameters
Decay in Terms of Half-Life	$N (t) = N_{0} (\frac{1}{2})^{t / T_{1/2}}$	Simplifies calculation for whole-number half-lives
Decay Q-Value	$Q = (m_{parent} - \sum m_{products}) c^{2}$	$Q > 0$ = spontaneous decay; $1 u \cdot c^{2} = 931.5 MeV$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Half-life activity calculation
2022 · FRQ
Balancing decay equations
2021 · MCQ
Spontaneous decay Q-value

What's Next

Nuclear decay is the foundational prerequisite for all other nuclear physics topics in AP Physics 2 Unit 7. Next, you will apply the concepts of spontaneous decay, mass-energy conversion, and half-life to binding energy per nucleon, nuclear fission and fusion, and radiometric dating applications. Without mastering conservation laws for decay, half-life calculations, and Q-value analysis, you will not be able to correctly analyze fission reactions or calculate the energy released in fusion, which are common high-weight FRQ topics on the AP exam. This topic also connects to earlier atomic physics concepts through gamma decay, where excited nuclear energy levels emit high-energy gamma photons analogous to atomic photon emission from excited electron states.

→Mass-Energy Equivalence