Nuclear Excited States and Gamma Emission Revision Notes | A Level AQA Physics

1. Why Nuclei Get Excited — and What Happens Next

After an alpha or beta decay, the daughter nucleus does not always land in its lowest-energy configuration. Think of it like a ball rolling down a staircase: it might stop on a middle step rather than reaching the floor. That middle step is the nuclear excited state.

The nucleus holds surplus energy above its ground state. To shed that energy, it emits a gamma ray — a high-energy photon of electromagnetic radiation. This transition from excited state to ground state is gamma emission.

A crucial point: the excited nucleus is still the same element with the same nucleon number. The proton number $Z$ and mass number $A$ do not change during gamma emission — only energy is released. This distinguishes gamma emission sharply from alpha and beta decay, both of which change the nuclear composition.

2. The Mechanism of Gamma Emission

Picture the nucleus as a system with discrete energy levels, analogous to the electron energy levels you met in atomic physics. When a nucleus drops from an excited level $E^*$ to the ground state $E_0$, the energy difference $\Delta E = E^* - E_0$ is carried away by a single gamma photon.

Because nuclear energy-level gaps are far larger than atomic electron-level gaps, gamma photons typically carry more energy than visible light and overlap with or exceed X-ray energies. The key distinction between gamma rays and X-rays is their origin — gamma rays are produced by nuclear transitions, whereas X-rays arise from electron transitions or deceleration — rather than energy alone. Typical gamma energies lie in the range of tens of keV to several MeV, corresponding to wavelengths shorter than $10^{-11}$ m.

The sequence of events is worth fixing clearly in your mind. First, an alpha or beta decay produces a daughter nucleus in an excited state. Then, almost instantaneously (often within $10^{-12}$ s), the excited nucleus emits a gamma photon and drops to its ground state. The two events — the initial decay and the gamma emission — are distinct, even though they happen in rapid succession.

3. Representing Gamma Emission — Notation and Nuclear Equations

Gamma rays are represented in nuclear equations as $^{0}_{0}\gamma$. Both the mass number and the atomic number are zero, confirming that no nucleons are transferred and no charge is carried away. The equation below shows a nucleus $X$ in an excited state (denoted by an asterisk) decaying to its ground state:

A concrete example: cobalt-60 undergoes beta-minus decay to produce nickel-60 in an excited state. The excited nickel-60 nucleus then emits two gamma photons in rapid succession as it cascades down to its ground state. The beta decay and the gamma emission are written as separate equations.

4. Metastable States — When the Excited State Lingers

Most nuclear excited states decay within picoseconds. Occasionally, however, a nucleus gets 'stuck' in an excited state for a measurable period — seconds, hours, or even longer. These long-lived excited states are called metastable isomers, and the 'm' suffix in a nuclide symbol flags this condition.

The metastable state is not a different element — it is the same nucleus with the same proton and nucleon numbers, carrying excess energy. When it finally emits its gamma photon, it becomes the ground-state nucleus of the same nuclide. This is why technetium-99m decays to technetium-99, not to a different element.

5. Technetium-99m — A Gamma Source in Medical Diagnosis

Technetium-99m ($^{99m}_{43}\text{Tc}$) is the most widely used radioisotope in nuclear medicine. Its properties align almost perfectly with what a medical imaging source needs to be: detectable outside the body, minimally harmful to the patient, and short-lived enough to clear quickly.

The gamma emission from $^{99m}\text{Tc}$ has an energy of approximately 140 keV. This energy is high enough to escape the body and be detected by a gamma camera, yet low enough to be absorbed efficiently by the detector rather than passing straight through it. Lower-energy photons would be absorbed by tissue before reaching the camera; higher-energy photons would pass through the detector without being registered.

The 6-hour half-life is a carefully balanced compromise. After a typical scan lasting 1–2 hours, the activity in the patient falls rapidly: after 24 hours (four half-lives), the activity has dropped to $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ of the administered value. This minimises the radiation dose received by the patient while still allowing the imaging procedure to be completed.

Worked example: Calculating Remaining Activity of Tc-99m After Imaging

A patient is administered a sample of $^{99m}\text{Tc}$ with an initial activity of $800$ MBq. The half-life of $^{99m}\text{Tc}$ is 6.0 hours. Calculate the activity remaining after 24 hours.

1. $n = \frac{t}{t_{1/2}} = \frac{24 \text{ h}}{6.0 \text{ h}} = 4$ — Divide total time by the half-life. Here 24 hours contains exactly 4 half-lives. [M1]
2. $A = A_0 \times \left(\frac{1}{2}\right)^n = 800 \times \left(\frac{1}{2}\right)^4$ — Each half-life halves the activity. After n half-lives, the fraction remaining is $(1/2)^n$. [M1]
3. $\left(\frac{1}{2}\right)^4 = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}$ — Expanding the power step by step: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$, then $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$, then $\frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$.
4. $A = 800 \times \frac{1}{16} = 50 \text{ MBq}$ — $800 \div 16 = 50$. The activity has fallen to 50 MBq after 24 hours. [A1]
5. 4 halvings: $800 \to 400 \to 200 \to 100 \to 50$ MBq — Tracing each halving individually confirms the answer. This is a reliable cross-check for half-life calculations.

Answer: $A = 50$ MBq after 24 hours

In practice, $^{99m}\text{Tc}$ is produced in a hospital from a molybdenum-99 generator. Molybdenum-99 undergoes beta-minus decay with a half-life of about 66 hours to produce $^{99m}\text{Tc}$, which is then chemically extracted and attached to a targeting molecule before injection. The relatively long half-life of the parent Mo-99 means the generator can supply fresh Tc-99m over several days.

6. Exam Technique — Getting Full Marks on Gamma and Tc-99m Questions

AQA questions on this topic tend to fall into three categories: writing or completing nuclear equations, explaining why gamma emission does not change the nucleus, and evaluating the suitability of Tc-99m for medical use. Each type has its own mark-winning language.

For nuclear equations, always check that both the mass numbers and the proton numbers balance on each side. A common slip is to write the gamma symbol without the superscript and subscript zeros — examiners expect $^{0}_{0}\gamma$, not just $\gamma$, to confirm you understand no nucleons are transferred.

For 'explain' questions about gamma emission, the mark scheme typically wants two linked ideas: the nucleus is in an excited state (above ground state energy), and it releases the excess energy as a gamma photon when it transitions to the ground state. Stating only that 'the nucleus emits a gamma ray' without explaining the energy-level transition will not score the explanation mark.