Nuclear Decay — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: alpha, beta, and gamma decay types, standard nuclear notation, conservation of charge and nucleon number, the exponential decay law, activity, half-life calculations, and mass-energy equivalence for spontaneous decay reactions.

You should already know: Basic atomic structure (protons, neutrons, isotopes), the mass-energy relation $E=m{c}^2$, and fundamental exponential function properties.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Nuclear Decay?

Nuclear decay is the spontaneous breakdown of an unstable atomic nucleus, where the nucleus emits ionizing radiation to transition to a more stable lower-energy configuration. Nuclei become unstable when they have an incorrect neutron-to-proton ratio, too large a total nucleon count, or exist in an excited energy state after a previous nuclear reaction. For AP Physics 2 CED, this topic accounts for approximately 2-3% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with applications to radiometric dating or medical nuclear physics. Standard notation for a nuclide follows the convention ${}_Z^A\text{X}$, where $A$ = mass (nucleon) number (total protons + neutrons), $Z$ = atomic (proton/charge) number, and X is the element’s chemical symbol. Common synonyms include radioactive decay and spontaneous nuclear transmutation (a term used specifically when decay changes the element identity, which occurs for alpha and beta decay, but not gamma decay). Because decay is a quantum process, it is impossible to predict when any single nucleus will decay; however, large collections of unstable nuclei follow predictable statistical behavior that gives rise to the exponential decay law.

2. Types of Nuclear Decay and Conservation Laws

All nuclear decay reactions must satisfy two fundamental conservation laws consistently tested on the AP Physics 2 exam: conservation of total nucleon number ($A$) and conservation of total electric charge (proton number $Z$). This means the sum of $A$ values for all reactants (the parent nuclide) equals the sum of $A$ values for all products (the daughter nuclide + emitted particles), and the same equality holds for $Z$ values. There are three primary decay types you must know:

1. Alpha decay: Occurs for very heavy nuclides ($A>200$) that are too large to be stable. The parent emits an alpha particle, which is identical to a helium nucleus: ${}_2^4\alpha$ (or ${}_2^4\text{He}$). The general balanced form follows conservation: ${}_Z^A\text{X}{\to }_{Z-2}^{A-4}\text{Y}{+}_2^4\alpha$.
2. Beta-minus decay: The most common beta decay type, occurring when a nucleus has too many neutrons (too high a neutron-to-proton ratio). A neutron decays into a proton, a high-energy electron (${}_{-1}^0{\beta }^{-}$), and an antineutrino. The general form is: ${}_Z^A\text{X}{\to }_{Z+1}^A\text{Y}{+}_{-1}^0{\beta }^{-}+\bar{\nu }$. Total nucleon number stays the same (1 neutron becomes 1 proton), while Z increases by 1 to conserve charge.
3. Gamma decay: Occurs when a nucleus is left in an excited energy state after a previous decay or reaction. It 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

Problem: Carbon-14 (${}_6^{14}\text{C}$) 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.

Solution:

1. Start with the general beta-minus decay skeleton: ${}_6^{14}\text{C}{\to }_Z^A\text{N}{+}_{-1}^0{\beta }^{-}+\bar{\nu }$
2. Apply conservation of nucleon number: $14=A+0⟹A=14$
3. Apply conservation of charge number: $6=Z+(-1)⟹Z=6+1=7$
4. Write the full balanced equation: ${}_6^{14}\text{C}{\to }_7^{14}\text{N}{+}_{-1}^0{\beta }^{-}+\bar{\nu }$

Result: The daughter nuclide is nitrogen-14, with mass number 14 and atomic number 7.

Exam tip: Always confirm the 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.

3. Exponential Decay, Half-Life, and Activity

Radioactive decay is a statistical process: the instantaneous rate of decay is proportional to the number of undecayed nuclei remaining at that time, $N(t)$. This relationship gives the differential decay law: $\frac{dN}{dt}=-\lambda N$, where $\lambda$ is the decay constant (units of inverse time; a larger $\lambda$ means a faster decay rate). Solving this differential equation gives the exponential decay law for undecayed nuclei: $$N(t)=N_0{e}^{-\lambda t}$$ 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. When $t=T_{1/2}$, $N=N_0/2$. Substituting into the decay law and solving 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, equal to the number of decays per unit time: $R=|\frac{dN}{dt}|=\lambda N(t)$. 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 $N(t)$ and $R(t)$ 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

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

Solution:

1. Calculate the number of half-lives passed: $n=\frac{t}{T_{1/2}}=\frac{24}{8}=3$ half-lives.
2. Use the half-life activity formula: $R=R_0{\left(\frac{1}{2}\right)}^n=1280\times {\left(\frac{1}{2}\right)}^3=1280\times \frac{1}{8}=160\text{ MBq}$.
3. For the initial number of nuclei, use $R_0=\lambda N_0$. Convert $R_0$ to decays per day to match units: $1280\text{ MBq}=1.28\times 10^9\text{ decays/s}=1.28\times 10^9\times 86400=1.10592\times 10^{14}\text{ decays/day}$.
4. Solve for $N_0$: $N_0=\frac{R_0}{\lambda }=\frac{1.10592\times 10^{14}}{0.0866}\approx 1.28\times 10^{15}$ nuclei.

Exam 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.

4. Mass-Energy 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 difference in mass ($\Delta m$) is converted to kinetic energy of the decay products, per mass-energy equivalence $E=\Delta m{c}^2$. 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 and requires an input of energy. A convenient shortcut for AP problems is to use atomic masses (not nuclear masses) for Q-value calculations: the total number of electrons is the same on both sides of the decay equation for alpha and beta-minus decay, so electron masses cancel out automatically. Recall that 1 atomic mass unit (u) is equivalent to $931.5\text{ MeV}$ of energy: $1\text{ u}\cdot c^2=931.5\text{ MeV}$.

Worked Example

Problem: Polonium-210 (${}_{84}^{210}\text{Po}$, atomic mass = 209.98287 u) undergoes alpha decay to form lead-206 (${}_{82}^{206}\text{Pb}$, atomic mass = 205.97447 u) and an alpha particle (${}_2^4\text{He}$, atomic mass = 4.00260 u). Calculate the Q-value of the decay and confirm it is spontaneous.

Solution:

1. Confirm the decay equation: ${}_{84}^{210}\text{Po}{\to }_{82}^{206}\text{Pb}{+}_2^4\text{He}$. Total electrons on left = 84, total electrons on right = 82 + 2 = 84, so electron masses cancel.
2. Calculate total product mass: $m_{\text{products}}=205.97447+4.00260=209.97707\text{ u}$.
3. Calculate mass difference: $\Delta m=m_{\text{parent}}-m_{\text{products}}=209.98287-209.97707=0.00580\text{ u}$.
4. Calculate Q-value: $Q=0.00580\text{ u}\cdot c^2\times 931.5\text{ MeV}/(\text{u}\cdot c^2)\approx 5.40\text{ MeV}$.

Since $Q=+5.40\text{ MeV}>0$, the decay is spontaneous, as expected.

Exam 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 the tabulated atomic masses directly.

5. Common Pitfalls (and how to avoid them)

• 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 $\lambda =\frac{T_{1/2}}{\ln 2}$ instead of $T_{1/2}=\frac{\ln 2}{\lambda }$, leading to a factor of ~2 error in half-life or decay constant calculations. Why: Students misremember which quantity goes in the numerator. 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 $Q=(\sum m_{\text{products}}-m_{\text{parent}})c^2$, 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 energy.
• 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 ${}_0^0\gamma$, so A and Z of the daughter are identical to the parent.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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

Worked Solution: First calculate how many half-lives have passed: $n=60/15=4$ half-lives. Activity after $n$ half-lives is $R=R_0(1/2{)}^n=800\ast (1/2{)}^4=800/16=50$ Bq. The other options correspond to incorrect values of n: 100 Bq is for 3 half-lives, 200 Bq for 2, 400 Bq for 1. The correct answer is C.

Question 2 (Free Response)

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. (a) Calculate the decay constant of potassium-40. (b) How many half-lives have passed since the rock formed? (c) What is the age of the rock?

Worked Solution: (a) Use the half-life-decay constant relation: $\lambda =\frac{\ln 2}{T_{1/2}}=\frac{0.693}{1.25\times 10^9\text{ years}}\approx 5.54\times 10^{-10}{\text{ year}}^{-1}$. (b) The ratio of K-40 to Ar-40 is 1:3, so 1 part undecayed K-40, 3 parts decayed Ar-40. That means 1/4 of the original K-40 remains: $N/N_0=1/4=(1/2{)}^n$, so $n=2$ half-lives. (c) Age = $n{T}_{1/2}=2\ast 1.25\times 10^9=2.5\times 10^9$ years, or 2.5 billion years.

Question 3 (Application / Real-World Style)

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 for the scan, what is the activity when it is used? Does the sample meet the minimum required activity of 25 MBq for the scan?

Worked Solution: Number of half-lives passed: $n=24/6=4$ half-lives. Activity when used: $R=500\text{ MBq}\ast (1/2{)}^4=500/16=31.25\text{ MBq}$. The activity when the sample is used is ~31 MBq, which is above the minimum required 25 MBq, so the sample is still usable for the scan. The 24-hour waiting period reduced the activity by a factor of 16, but the remaining activity is sufficient for the imaging procedure.

7. Quick Reference Cheatsheet

8. What's Next

Nuclear decay is the foundational prerequisite for all other nuclear physics topics you will study next 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 (energy level transitions, photon emission) through gamma decay, where excited nuclear energy levels emit high-energy gamma photons analogous to atomic photon emission from excited electron states.

• Nuclear Binding Energy and Mass Defect
• Nuclear Fission and Fusion
• Applications of Nuclear Physics
• Photon Energy and Atomic Spectra