A LevelPhysicsAQARevision NotesNuclear PhysicsNuclear Instability & RadiusNuclear Excited States

Nuclear Excited States (AQA A Level Physics): Revision Note

Exam code: 7408

Written by: Katie M

Reviewed by: Tim

Updated on 6 October 2026

Guided study available on this topic

Understand this topic with deeper explanations and understanding checks.

Nuclear excited states

A nucleus can exist in an excited state in the same way as an electron
Once an unstable nucleus has decayed, it may emit any remaining energy in the form of a gamma photon ( $γ$ )
- The emission of a $γ$ photon does not change the number of protons or neutrons in the nucleus
- It only allows the nucleus to lose energy

This happens when a daughter nucleus is in an excited state after a decay
This excited state is usually very short-lived, and the nucleus quickly moves to its ground state, either directly or via one or more lower-energy excited states

Diagram showing uranium-235 absorbing a neutron and fissioning into molybdenum-99 plus another fission product. Molybdenum-99 then decays to metastable technetium-99m. — **Metastable technetium-99m is a technetium-99 nucleus in an excited energy state**

One common application of this is the use of technetium-99m as a $γ$ source in medical diagnosis
- The ‘m’ stands for metastable which means the nucleus exists in a particularly stable excited state

Technetium-99m is the decay product of molybdenum-99, which can be found as a product in nuclear reactors
The decay of molybdenum-99 is shown below:

${}_{42}^{99}{Mo} \to {}_{43}^{99 m}{Tc} + {}_{- 1}^{0}β + v_{e}$

${}_{43}^{99 m}{Tc} \to {}_{43}^{99}{Tc} + γ$

The half-life of molybdenum-99 is 66 hours
- This is long enough for the sample to be transported to hospitals
- Subsequently, the technetium-99m can be separated at the hospital

Technetium-99m has a short half-life of 6 hours
- This is an adequate timeframe for examining a patient
- Plus, it is short enough to minimise damage to the patient

Nuclear energy level diagrams

Nuclear energy levels are similar to electron energy levels
The nuclear energy level diagram of molybdenum-99 can be represented as follows:

The decay mode (usually alpha or beta) is represented by a diagonal line
The excited state, or states, are generally stacked in descending energy order to the right of the decay

Worked Example

A nucleus of iron Fe-59 decays into a stable nucleus of cobalt Co-59. It decays by $β^{-}$ emission followed by the emission of $γ$ -radiation as the Co-59 nucleus de-excites into its ground state.

The total energy released when the Fe-59 nucleus decays is 2.52 × 10^–13 J.

The Fe-59 nucleus can decay to one of three excited states of the cobalt-59 nucleus as shown below. The energies of the excited states are shown relative to the ground state.

Energy-level diagram of Fe-59 beta-minus decay to three Co-59 excited states at 2.29 times 10 to the minus 13 joules, 2.06 times 10 to the minus 13 joules and 1.76 times 10 to the minus 13 joules. Each transition emits a beta-minus particle and antineutrino. Co-59 ground state is at 0 joules.

Following the production of excited states of Co-59, $γ$ -radiation of discrete wavelengths is emitted.

(a) Calculate the maximum possible kinetic energy of the $β^{-}$ particle emitted in MeV. [2]

(b) State the maximum number of discrete wavelengths that could be emitted. [1]

Answer:

Part (a)

Step 1: Identify the beta emission with the largest energy gap

Energy-level diagram showing Fe-59 at 2.52 × 10⁻¹³ joules decaying to Co-59 excited states. The blue arrow highlights the largest energy gap, to the 1.76 × 10⁻¹³ joule state, via beta-minus emission and an antineutrino.

[1 mark]

Step 2: Calculate the energy difference

$∆ E = 2.52 \times 10^{- 13} - 1.76 \times 10^{- 13} = 7.6 \times 10^{- 14} J$

Step 3: Convert from J to MeV

1 MeV = 1.6 × 10^–13 J

$∆ E = \frac{7.6 \times 10^{- 14}}{1.6 \times 10^{- 13}} = 0.475 MeV$ [1 mark]

Part (b)

Energy-level diagram showing Fe-59 beta-minus decay to Co-59 levels at 2.29, 2.06 and 1.76 times 10 to the minus 13 joules, with six possible gamma transitions between the three excited states and ground state.

There are six possible transitions, hence six discrete wavelengths could be emitted [1 mark]

Part (c)

Step 1: Identify the emission with the longest wavelength / smallest energy gap

Energy-level diagram showing Fe-59 beta-minus decay to Co-59 states at 2.29, 2.06 and 1.76 times 10 to the minus 13 joules. A purple arrow highlights the 2.29 times 10 to the minus 13 joules to 2.06 times 10 to the minus 13 joules transition.

Longest wavelength = lowest frequency = smallest energy [1 mark]

Step 2: Calculate the energy difference

$∆ E = 2.29 \times 10^{- 13} - 2.06 \times 10^{- 13} = 2.3 \times 10^{- 14} J$ [1 mark]

Step 3: Write down de Broglie’s wavelength equation

$E = \frac{h c}{λ}$

Where:

$h$ = Planck’s constant
$c$ = speed of light

Step 4: Calculate the wavelength associated with the energy change

$λ = \frac{h c}{E} = \frac{(6.63 \times 10^{- 34}) \times (3.0 \times 10^{8})}{2.3 \times 10^{- 14}}$

$λ = 8.65 \times 10^{- 12} m (3 s . f .)$ [1 mark]

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