Nuclear Energy Levels (DP IB Physics: HL): Revision Note

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.

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

(b) State the maximum number of discrete wavelengths that could be emitted

(c) Calculate the longest wavelength of the emitted γ-radiation

Answer:

Part (a)

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

Step 2: Calculate the energy difference

$∆E$ = (2.52 – 1.76) × 10^–13 = 7.6 × 10^–14 J

Step 3: Convert from J to MeV

$∆E=\frac{7.6\times {10}^{-14}}{1.6\times {10}^{-13}}=0.475=0.48\mathrm{MeV}$

• 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

$∆E$ = (2.29 – 2.06) × 10^–13 = 2.3 × 10^–14 J

Step 3: Relate photon energy and wavelength

$E=\frac{hc}{\lambda }$

• Where:

  • h = Planck’s constant

  • c = speed of light

Step 4: Calculate the wavelength associated with the energy change

$\lambda =\frac{hc}{E}=\frac{\left(6.63\times {10}^{-34}\right)\left(3.0\times {10}^8\right)}{2.3\times {10}^{-14}}=8.6\times {10}^{12}\mathrm{m}$