E3-2. Decay Equations, Mass Defect, and Binding Energy HL

1. Radioactive Decay Equations

Conservation Laws in Nuclear Decay

  • During any nuclear reaction or radioactive decay, specific fundamental quantities must be strictly conserved. The sum of these values on the left side of the equation must equal the sum on the right side.
  • Conservation of Nucleon Number ($A$): The total number of protons and neutrons remains constant.
  • Conservation of Charge ($Z$): The total fundamental charge (represented by the atomic number) remains constant.
  • Conservation of Mass-Energy: The total energy (including the energy equivalent of rest mass) is conserved.

Transmutation Examples

  • Alpha Decay: An unstable heavy nucleus emits a helium nucleus. $A$ decreases by 4, $Z$ decreases by 2. $${}_{92}^{238}\text{U} \longrightarrow {}_{90}^{234}\text{Th} + {}_{2}^{4}\text{He}$$
  • Beta Minus Decay ($\beta^-$): A neutron converts into a proton, an electron, and an anti-neutrino. $A$ remains constant, $Z$ increases by 1. $${}_{6}^{14}\text{C} \longrightarrow {}_{7}^{14}\text{N} + {}_{-1}^{0}\text{e} + \bar{\nu}_e$$

2. Mass Defect and Nuclear Binding Energy

The Mass Defect ($\Delta m$)

  • If you measure the mass of a stable nucleus, it is shockingly less than the sum of the masses of its individual, separated constituent protons and neutrons.
  • This difference in mass is called the mass defect. It represents the mass that was completely converted into energy and radiated away when the nucleus first formed.
  • $$\Delta m = (Z \cdot m_{\text{proton}} + N \cdot m_{\text{neutron}}) - m_{\text{nucleus}}$$

Einstein's Mass-Energy Equivalence and Binding Energy

  • The energy corresponding to this missing mass is called the Nuclear Binding Energy. It is the absolute minimum energy that must be supplied to completely dismantle a nucleus into its constituent nucleons.
  • It is calculated using Einstein's famous equivalence principle:
    $$E = \Delta m c^2$$
  • In nuclear physics, it is heavily preferred to use the atomic mass unit ($u$) rather than kilograms. $1\text{ u}$ is defined as exactly $1/12$ the mass of a Carbon-12 atom.
  • Using $E=mc^2$, $1\text{ u}$ of mass is equivalent to exactly $931.5\text{ MeV}$ of energy. Therefore: $$E (\text{in MeV}) = \Delta m (\text{in u}) \times 931.5$$

3. Binding Energy per Nucleon Curve

Comparing the total binding energy of different nuclei is misleading because larger nuclei naturally have more binding energy simply by having more nucleons. Instead, physicists map the Binding Energy per Nucleon ($BE/A$) against the mass number $A$.

Binding Energy per Nucleon Curve Average BE per Nucleon (MeV) 0 2 4 6 8 10 Mass Number (A) 0 50 100 150 200 250 Fe-56 (Peak ~8.8 MeV) U-235 H-2 FUSION FISSION
  • The higher the $BE/A$, the more tightly bound and stable the nucleus is.
  • The curve rises sharply for light elements, reaches a global maximum at Iron-56 (${}^{56}\text{Fe}$) (which has $BE/A \approx 8.8\text{ MeV}$), and then slowly drops off for heavier elements like Uranium.
  • Nuclear Fusion: Light nuclei ($A < 56$) can fuse together to form a heavier nucleus. The new nucleus has a higher $BE/A$, meaning it is more stable, and the excess mass-energy is released.
  • Nuclear Fission: Heavy nuclei ($A > 56$) can split into lighter fragments. The fragments possess a higher $BE/A$ than the heavy parent, again resulting in a massive release of energy.

4. Advanced HL Examples

Example 1: Calculating Binding Energy per Nucleon

Problem: Determine the binding energy per nucleon for an atom of Iron-56 (${}^{56}_{26}\text{Fe}$).
Given: Mass of a proton = $1.00728\text{ u}$, Mass of a neutron = $1.00867\text{ u}$, Mass of an entire Iron-56 nucleus = $55.92067\text{ u}$, $1\text{ u} = 931.5\text{ MeV}$.

Solution:

Step 1: Calculate the expected mass of the separated nucleons.
Iron-56 has $Z = 26$ protons and $N = 56 - 26 = 30$ neutrons. $$\text{Expected Mass} = (26 \times 1.00728) + (30 \times 1.00867) = 26.18928 + 30.26010 = 56.44938\text{ u}$$

Step 2: Calculate the Mass Defect ($\Delta m$).
$$\Delta m = 56.44938\text{ u} - 55.92067\text{ u} = 0.52871\text{ u}$$

Step 3: Calculate total Binding Energy (BE) and Binding Energy per Nucleon (BE/A).
$$\text{Total BE} = 0.52871 \times 931.5\text{ MeV/u} = 492.49\text{ MeV}$$ $$\text{BE/A} = \frac{492.49\text{ MeV}}{56\text{ nucleons}} = \mathbf{8.79\text{ MeV / nucleon}}$$ This perfectly aligns with the peak of the binding energy curve, proving Iron is the most stable element in the universe.

Example 2: The $Q$-Value of Alpha Decay and Recoil Kinematics

Problem: Radium-226 undergoes alpha decay to become Radon-222.
Masses: ${}^{226}\text{Ra} = 226.0254\text{ u}$; ${}^{222}\text{Rn} = 222.0176\text{ u}$; ${}^{4}\text{He} = 4.0026\text{ u}$.
(a) Calculate the $Q$-value (total energy released) of the reaction in MeV.
(b) Assuming the Radium nucleus was initially at rest, calculate the exact kinetic energy of the emitted alpha particle, accounting for the recoil of the Radon nucleus.

Solution:

Part (a): Mass Defect and $Q$-value:
The $Q$-value represents the energy equivalent of the mass lost during the decay. $$\Delta m = m_{\text{Reactants}} - m_{\text{Products}}$$ $$\Delta m = 226.0254\text{ u} - (222.0176\text{ u} + 4.0026\text{ u}) = 226.0254 - 226.0202 = 0.0052\text{ u}$$ $$Q = 0.0052\text{ u} \times 931.5\text{ MeV/u} = \mathbf{4.84\text{ MeV}}$$

Part (b): Two-body kinematics (Kinetic energy distribution):
The total energy released ($Q$) must be distributed as kinetic energy between the alpha particle and the recoiling Radon nucleus. By conservation of momentum ($p_{\alpha} = p_{\text{Rn}}$), the lighter alpha particle takes the vast majority of the kinetic energy ($K = p^2/2m$). The specific fraction of $Q$ taken by the alpha particle is given by the mass ratio: $$K_{\alpha} = \left( \frac{m_{\text{Rn}}}{m_{\text{Rn}} + m_{\alpha}} \right) Q$$ $$K_{\alpha} = \left( \frac{222}{222 + 4} \right) \times 4.84\text{ MeV} = \left( \frac{222}{226} \right) \times 4.84 \approx \mathbf{4.75\text{ MeV}}$$ The remaining $0.09\text{ MeV}$ is the recoil kinetic energy of the heavy Radon nucleus.

Example 3: Fusion Energetics & Mass Energy Density

Problem: In a future fusion reactor, Deuterium (${}^{2}_{1}\text{H}$) and Tritium (${}^{3}_{1}\text{H}$) fuse to form Helium-4 and a free neutron. The $Q$-value of this reaction is $17.6\text{ MeV}$. Calculate the total energy released in Joules if $1.0\text{ kg}$ of a 50/50 molar mix of Deuterium and Tritium is completely fused. Compare this to the combustion of $1\text{ kg}$ of coal ($24\text{ MJ}$).

Solution:

Step 1: Convert $Q$-value per reaction to Joules. $$Q = 17.6 \times 10^6\text{ eV} \times (1.60 \times 10^{-19}\text{ J/eV}) = 2.816 \times 10^{-12}\text{ J per reaction}$$

Step 2: Find the number of reactions in $1\text{ kg}$ of fuel.
The "molar mass" of one reaction pair (1 D + 1 T) is approximately $2\text{ g/mol} + 3\text{ g/mol} = 5\text{ g/mol} = 0.005\text{ kg/mol}$.
Number of moles of pairs in $1\text{ kg}$: $n = \frac{1.0\text{ kg}}{0.005\text{ kg/mol}} = 200\text{ moles}$.
Number of individual reactions ($N$) = $200 \times N_A = 200 \times (6.02 \times 10^{23}) = 1.204 \times 10^{26}\text{ reactions}$.

Step 3: Total Energy Released. $$E_{\text{total}} = N \times Q = (1.204 \times 10^{26}) \times (2.816 \times 10^{-12}\text{ J}) = \mathbf{3.39 \times 10^{14}\text{ J}}$$ Comparison: $3.39 \times 10^{14}\text{ J}$ vs $2.4 \times 10^7\text{ J}$ for coal. $1\text{ kg}$ of fusion fuel produces over 14 million times more energy than $1\text{ kg}$ of coal!