AQA A Level Physics复习笔记8.3.3 Nuclear Excited States

• In the same way that electrons can exist in excited states, nuclei can also exist in excited states
• After an unstable nucleus emits an alpha particle, beta particle or undergoes electron capture, it may emit any remaining energy in the form of a gamma photon (γ)
  • 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

• 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:

• 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 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

Part (a)

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

Step 2: Calculate the energy difference

ΔE = (2.52 – 1.76) × 10–13 = 7.6 × 10–14 J

Step 3: Convert from J to MeV

• • 1 MeV = 1.6 × 10–13 J

Part (b)

• • There are 6 possible transitions, hence 6 discrete wavelengths could be emitted

Part (c)

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

• • Longest wavelength = lowest frequency = smallest energy

Step 2: Calculate the energy difference

ΔE = (2.29 – 2.06) × 10–13 = 2.3 × 10–14 J

Step 3: Write down de Broglie’s wavelength equation

Where:

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

   

Step 4: Calculate the wavelength associated with the energy change