E3-4. Nuclear Stability, Energy Levels & The Neutrino HL
1. Nuclear Stability and the Segrè Chart
The stability of a nucleus depends almost entirely on its ratio of neutrons to protons ($N/Z$). If we plot all known stable isotopes on a graph of Neutron Number ($N$) vs. Proton Number ($Z$), they form a distinct "Band of Stability".
- Low Mass Nuclei ($Z < 20$): Stable nuclei have an $N/Z$ ratio of approximately 1:1. The strong nuclear force perfectly balances the electrostatic proton repulsion.
- Heavy Nuclei ($Z > 20$): As the nucleus gets larger, electrostatic repulsion between protons operates over long distances, while the attractive strong nuclear force only operates over very short distances. To maintain stability without adding more repulsion, heavy nuclei must accumulate more neutrons. The stable $N/Z$ ratio curves upwards, reaching about 1.5:1 for Lead-208.
- Nuclei falling above the stability band have too many neutrons. They will undergo Beta Minus ($\beta^-$) decay (converting a neutron to a proton) to move down and right towards the band.
- Nuclei falling below the stability band have too many protons. They will undergo Beta Plus ($\beta^+$) decay (converting a proton to a neutron) to move up and left towards the band.
2. Nuclear Energy Levels and Gamma Decay
Much like atomic electrons, the nucleons (protons and neutrons) inside a nucleus exist in discrete, quantized energy levels.
- When a nucleus undergoes alpha or beta decay, the "daughter" nucleus is frequently left in an excited, high-energy structural state (indicated by a $^*$ symbol, e.g., ${}^{60}\text{Ni}^*$).
- To reach the lowest-energy ground state, the nucleons shift their configuration, shedding the excess energy as a single, extremely high-frequency Gamma ($\gamma$) ray photon.
- Because the nuclear energy levels are completely discrete, the emitted gamma rays have exact, specific energies (producing a discrete line spectrum), similar to atomic emission spectra but at millions of times higher energies (MeV instead of eV).
3. Evidence for the Neutrino (Beta Decay Spectrum)
During alpha decay, alpha particles are emitted with very specific, discrete kinetic energies. However, early 20th-century measurements of beta decay revealed something shocking: beta particles were emitted with a continuous range of kinetic energies, up to a maximum end-point limit.
- Because energy states in the nucleus are discrete, the total energy released in a beta decay ($Q$-value) is a fixed constant.
- If beta decay only produced an electron ($X \rightarrow Y + e^-$), the electron *must* carry away that exact fixed energy to conserve momentum and energy. The continuous spectrum meant energy was seemingly vanishing, violating the First Law of Thermodynamics!
- To save the laws of physics, Wolfgang Pauli hypothesized the existence of a third, invisible particle emitted during the decay that shares the kinetic energy with the beta particle. Enrico Fermi later named it the neutrino.
- The neutrino has zero charge, practically zero mass, and interacts so weakly with matter it went undetected for decades. It perfectly accounts for the "missing" momentum and energy in the continuous beta spectrum.
4. Advanced HL Examples
Example 1: Predicting Transmutation Pathways
Problem: A nucleus of Carbon-14 (${}^{14}_{6}\text{C}$) is unstable. Using its position relative to the $N=Z$ line, predict its primary mode of decay, write the full nuclear equation including the correct neutrino species, and explain how its position on the Segrè chart changes.
Solution:
Carbon-14 has $Z = 6$ and $N = 14 - 6 = 8$. Its $N/Z$ ratio is $8/6 = 1.33$. For light elements ($Z < 20$), the stability line is at $N/Z \approx 1$. Because Carbon-14 sits well above the band of stability, it is neutron-rich.
To stabilize, it must convert a neutron into a proton via Beta Minus ($\beta^-$) decay. The equation is: $${}_{6}^{14}\text{C} \longrightarrow {}_{7}^{14}\text{N} + {}_{-1}^{0}\beta^- + \bar{\nu}_e$$ (Note the inclusion of the electron anti-neutrino, $\bar{\nu}_e$, which accompanies $\beta^-$ decay to conserve lepton number).
Chart Movement: The daughter nucleus Nitrogen-14 has $Z=7, N=7$ ($N/Z = 1$). On the $N-Z$ Segrè chart, the nucleus has moved down one unit in $N$ and right one unit in $Z$, successfully landing exactly on the stable $1:1$ line.
Example 2: Nuclear Energy Level Transitions (Gamma Decay)
Problem: Following a beta decay, a daughter nucleus is left in an excited state $E_2 = 1.33\text{ MeV}$ above the ground state ($E_0 = 0\text{ MeV}$). It transitions to an intermediate state $E_1 = 0.51\text{ MeV}$, and then finally drops to the ground state, emitting two distinct gamma ray photons in cascade. Calculate the exact wavelength (in picometers) of the highest frequency gamma ray emitted.
Solution:
Step 1: Determine the energy changes ($\Delta E$).
There are two transitions:
• $E_2 \rightarrow E_1$: $\Delta E = 1.33 - 0.51 = 0.82\text{ MeV}$
• $E_1 \rightarrow E_0$: $\Delta E = 0.51 - 0 = 0.51\text{ MeV}$
The highest frequency photon corresponds to the largest energy gap. Therefore, we use $\Delta E = 0.82\text{ MeV}$.
Step 2: Convert energy to Joules.
$$0.82\text{ MeV} = 0.82 \times 10^6\text{ eV} \times (1.60 \times 10^{-19}\text{ J/eV}) = 1.312 \times 10^{-13}\text{ J}$$
Step 3: Calculate the wavelength using $E = \frac{hc}{\lambda}$.
$$\lambda = \frac{hc}{\Delta E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1.312 \times 10^{-13}} = \mathbf{1.52 \times 10^{-12}\text{ m}}$$
Converting to picometers ($1\text{ pm} = 10^{-12}\text{ m}$), the wavelength is $1.52\text{ pm}$.
Example 3: 3-Body Kinematics and the Beta Endpoint
Problem: A nucleus undergoes beta decay, releasing a total $Q$-value energy of $1.17\text{ MeV}$. Draw a conclusion regarding the kinetic energy of the electron anti-neutrino when the emitted beta particle is detected moving with exactly $0.45\text{ MeV}$ of kinetic energy. What is the physical significance of the "end-point energy" on a beta spectrum graph?
Solution:
Because the heavy daughter nucleus experiences virtually zero recoil kinetic energy (due to its massive inertia relative to the electron), the total $Q$-value is shared almost entirely between the beta particle and the anti-neutrino. $$Q = K_{\beta} + K_{\nu}$$
If the beta particle is detected with $K_{\beta} = 0.45\text{ MeV}$, the anti-neutrino must carry the remainder: $$K_{\nu} = 1.17\text{ MeV} - 0.45\text{ MeV} = \mathbf{0.72\text{ MeV}}$$
The End-Point Energy: If an event occurs where the anti-neutrino is emitted with approximately zero kinetic energy, the beta particle receives 100% of the available decay energy ($K_{\beta} = 1.17\text{ MeV}$). This maximum possible kinetic energy is the end-point energy on the continuous beta spectrum. Its exact value corresponds perfectly to the total mass defect ($Q$-value) of the transition.