# Nuclear Decay

> AP Physics 2 · Unit 7: Quantum, Atomic, and Nuclear Physics
> Source: https://www.owlsprep.com/study/ap-physics-2-u7-nuclear-decay/

This guide covers core concepts of nuclear decay for AP Physics 2, including decay types, conservation laws, exponential decay, half-life calculations, and Q-value for spontaneous decay reactions.

**Prerequisites:** Basic atomic structure (protons, neutrons, isotopes); Mass-energy equivalence $E=mc^2$; Fundamental properties of exponential functions

## Learning objectives

- Write balanced nuclear decay equations for alpha, beta-minus, and gamma decay
- Calculate half-life, decay constant, and activity for radioactive samples
- Calculate the Q-value of a decay reaction and confirm spontaneous decay
- Apply exponential decay to solve real-world nuclear decay problems

## Nuclear Decay Fundamentals and Notation

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.

**Nuclide Notation** — 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 $\text{X}$ is the element's chemical symbol.

*Notation:* ^A_Z \text{X}

## Types of Decay and Conservation Laws

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 ($^4_2 \alpha$, 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 ($^0_{-1} \beta^-$), and an antineutrino, resulting in transmutation.
- **Gamma decay**: Occurs when a nucleus is in an excited energy state. Emits a high-energy gamma photon ($^0_0 \gamma$) to release excess energy, with no change to $A$ or $Z$, so no transmutation occurs.

**Worked example:** Carbon-14 ($^{14}_{6} \text{C}$) 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. $$^{14}_{6} \text{C} \rightarrow ^{14}_{7} \text{N} + ^0_{-1} \beta^- + \bar{\nu}$$

> **tip**
>
> Always confirm total charge and total nucleon number are equal on both sides of the decay equation. AP multiple-choice distractors almost always violate one of these conservation laws, so a 10-second check will eliminate wrong answers instantly.

## Exponential Decay, Half-Life, and Activity

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 $\frac{dN}{dt} = -\lambda N$, where $\lambda$ is the decay constant (units of inverse time, larger $\lambda$ 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 $R = \left|\frac{dN}{dt}\right| = \lambda N(t)$. Activity also follows the exponential decay law $R(t) = R_0 e^{-\lambda t}$, where $R_0 = \lambda N_0$ 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, $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$ and $R(t) = R_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.

**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 $\lambda = 0.0866 \text{ day}^{-1}$.

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 $R_0 = \lambda N_0$, convert activity to decays per day to match units of $\lambda$:
6. $$1280 \text{ MBq} = 1.10592 \times 10^{14} \text{ decays/day}$$
7. Solve for $N_0$:
8. $$N_0 = \frac{R_0}{\lambda} = \frac{1.10592 \times 10^{14}}{0.0866} \approx 1.28 \times 10^{15} \text{ nuclei}$$

> **tip**
>
> Always match units for $\lambda$ and activity: if $\lambda$ is given in inverse years, convert activity to decays per year, not per second. AP questions regularly mix units to test attention to detail.

## Mass-Energy Equivalence and Decay Q-Value

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 $\Delta m$ is converted to kinetic energy of the decay products, per $E = \Delta m c^2$. 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 $1 \text{ u} \cdot c^2 = 931.5 \text{ MeV}$.

**Worked example:** Polonium-210 ($^{210}_{84} \text{Po}$, atomic mass = 209.98287 u) undergoes alpha decay to form lead-206 ($^{206}_{82} \text{Pb}$, atomic mass = 205.97447 u) and an alpha particle ($^4_2 \text{He}$, 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.

> **tip**
>
> Don’t waste time subtracting electron masses when using atomic masses for alpha or beta-minus decay Q calculations: they always cancel out, so you can use tabulated atomic masses directly.

## AP Style Concept Check

**Check your understanding**

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

1. 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

   *Why:* 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.

2. A geologist is dating a volcanic rock sample that contains potassium-40, which has a half-life of $1.25 \times 10^9$ 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) $1.25 \times 10^9$ years
   - B) $2.5 \times 10^9$ years
   - C) $3.75 \times 10^9$ years
   - D) $5.0 \times 10^9$ years

   *Why:* 1/4 of the original potassium remains, which equals 2 half-lives, so age = $2 \times 1.25 \times 10^9 = 2.5 \times 10^9$ years.

3. 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?

   *Why:* 4 half-lives passed, so $R = 500 / 16 = 31.25$ MBq, which is above the 25 MBq minimum, so the sample is usable.

## Common pitfalls

- **Wrong:** 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 it fails: Students confuse the tiny mass of an electron with a nucleon, and mix up the sign of the beta particle’s charge.
  - Correct: Remember beta-minus decay converts a neutron to a proton, so total $A$ stays the same, and $Z$ increases by 1.
- **Wrong:** Using the inverse relation $\lambda = \frac{T_{1/2}}{\ln 2}$ instead of $T_{1/2} = \frac{\ln 2}{\lambda}$, leading to a factor of ~2 error in calculations.
  - Why it fails: Students misremember which quantity goes in the numerator of the half-life/decay constant relation.
  - Correct: Always verify your result with the $(1/2)^n$ half-life rule to cross-check for algebra errors.
- **Wrong:** Calculating Q-value as $Q = (\sum m_{\text{products}} - m_{\text{parent}}) c^2$, getting a negative Q even for spontaneous decay.
  - Why it fails: Students mix up the definition of Q-value with mass defect for binding energy.
  - Correct: Always write Q as (initial mass minus final mass) for decay, since the missing mass is converted to kinetic energy of products.
- **Wrong:** Claiming that after two half-lives, all original unstable nuclei have decayed.
  - Why it fails: Students misinterpret half-life as the total lifetime of the entire sample.
  - Correct: 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:** Changing the $A$ or $Z$ number of the parent nuclide for gamma decay.
  - Why it fails: Students assume all decay changes the nuclide identity, forgetting gamma is only energy emission.
  - Correct: Always write gamma as $^0_0 \gamma$, so $A$ and $Z$ of the daughter are identical to the parent.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Nuclide Notation | $^A_Z \text{X}$ | $A$ = total nucleon (mass) number, $Z$ = proton (charge) number |
| Conservation Laws | $\sum A_{\text{left}} = \sum A_{\text{right}}$, $\sum Z_{\text{left}} = \sum Z_{\text{right}}$ | Applies to all nuclear decay reactions |
| Alpha Decay General Form | $^A_Z \text{X} \rightarrow ^{A-4}_{Z-2} \text{Y} + ^4_2 \alpha$ | Occurs in heavy unstable nuclides |
| Beta-Minus Decay General Form | $^A_Z \text{X} \rightarrow ^A_{Z+1} \text{Y} + ^0_{-1} \beta^- + \bar{\nu}$ | Occurs in nuclides with too many neutrons |
| Exponential Decay Law | $N(t) = N_0 e^{-\lambda t}$, $R(t) = R_0 e^{-\lambda t}$ | $N$ = undecayed nuclei, $R$ = activity, $\lambda$ = decay constant |
| Half-Life / Decay Constant | $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$ | Relates the two common decay parameters |
| Decay in Terms of Half-Life | $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$ | Simplifies calculation for whole-number half-lives |
| Decay Q-Value | $Q = (m_{\text{parent}} - \sum m_{\text{products}}) c^2$ | $Q>0$ = spontaneous decay; $1 \text{ u} \cdot c^2 = 931.5 \text{ MeV}$ |

## 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](https://www.owlsprep.com/study/ap-physics-2-u7-mass-energy-equivalence/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-physics-2-u7-nuclear-decay/