Lecture 4: Nuclear Reactions Energy Balance and Charged Particle Interaction Physics

Nuclear Reaction Energy and Particle Energy Measurement

• The primary goal of studying nuclear reactions in this context is to calculate the energy involved, as radiation measurements specifically capture the kinetic energy of particles emitted during a reaction.

• Example reaction: A neutron is absorbed by Boron-10, which results in the emission of alpha particles.

• Measured Energy: Radiation measurement devices detect the energy of the alpha particle ($t_3$).

• Calibration: Calculating expected energy based on Q-values allows for the calibration of measurement tools to determine if experimental data is accurate or contains variances.

• Energy Variation: The energy of the emitted particle can vary based on the emission angle ($\theta$), such as $0^{\circ}$, $80^{\circ}$, or $90^{\circ}$.

Mathematical Framework for Nuclear Reactions

• Mass and Velocity Notation:

  • $m_1$: Particle 1 (Projectile, e.g., neutron).

  • $M_2$: Target nucleus (e.g., Boron).

  • $m_3$: Emitted particle (Product 1, e.g., alpha particle).

  • $M_4$: Recoil nucleus (Product 2).

• Assumptions for Reactants:

  • The target ($M_2$) is initially stationary ($v_2 = 0$), meaning its initial kinetic energy ($t_2$) is 0.

• Energy Balance Equation:

  • Every mass is expressed in terms of energy ($MeV$).

  • $m_1 + t_1 + m_2 = m_3 + t_3 + m_4$

  • Here, $t$ represents kinetic energy. In terms of classical mechanics, $t = \frac{1}{2} m v^2$.

• Momentum Balance (X-axis):

  • $m_1 v_1 = m_3 v_3 \cos(\theta) + m_4 v_4 \cos(\phi)$

  • The projectile travels horizontally. $\theta$ is the angle of the emitted particle relative to the projectile direction; $\phi$ is the angle of the recoil nucleus.

• Momentum Balance (Y-axis):

  • $0 = m_3 v_3 \sin(\theta) - m_4 v_4 \sin(\phi)$, or $m_3 v_3 \sin(\theta) = m_4 v_4 \sin(\phi)$.

• Rewriting in terms of Q-value:

  • $Q = m_1 + m_2 - m_3 - m_4$

  • Energy balance rewritten: $t_1 + Q = t_3 + t_4$

• Relating Momentum to Kinetic Energy:

  • Since $v = \sqrt{\frac{2t}{m}}$, the momentum equation for the X-axis becomes:

  • $\sqrt{2 m_1 t_1} = \sqrt{2 m_3 t_3} \cos(\theta) + \sqrt{2 m_4 t_4} \cos(\phi)$

Angular Cases for Particle Energy

• Case 1: Forward Emission ($\theta = 0^{\circ}$, $\phi = 180^{\circ}$)

  • The conservation equations simplify because $\cos(0) = 1$ and $\cos(180) = -1$.

  • $\sqrt{m_1 t_1} = \sqrt{m_3 t_3} - \sqrt{m_4 t_4}$

• Case 2: Right Angle Emission ($\theta = 90^{\circ}$)

  • The momentum vectors form a right-angled triangle where the resultant is $m_4 v_4$.

  • Using the Pythagorean theorem: $(m_1 v_1)^2 + (m_3 v_3)^2 = (m_4 v_4)^2$.

  • Substituting $t = \frac{1}{2} m v^2$: $m_1 t_1 + m_3 t_3 = m_4 t_4$.

• Determining $t_3$ (Emitted Particle Energy):

  • By combining the Pythagorean momentum relation with the energy balance ($t_1 + Q = t_3 + t_4$), the expression for $t_3$ at $90^{\circ}$ is:

  • $t_3 = \frac{(m_4 - m_1) t_1 + m_4 Q}{m_3 + m_4}$

Numerical Problem: Neutron Interaction with Nitrogen

• Reaction: ${}^1n + {}^{14}N \rightarrow \alpha + {}^{11}B$

• Given Data:

  • Neutron energy ($t_1$): $2\,MeV$.

  • Target Nitrogen is at rest.

  • Mass of ${}^{14}N$: $14.008074\,amu$

  • Mass of ${}^1n$: $1.008665\,amu$

  • Mass of Alpha ($\alpha$): $4.002604\,amu$

  • Mass of ${}^{11}B$: $11.009305\,amu$

• Q-Value Calculation:

  • $Q = (14.008074 + 1.008665) - (4.002604 + 11.009305) = -0.163\,MeV$

  • Note: Even though $Q$ is negative (endothermic), the reaction proceeds because the incident neutron is energetic ($2\,MeV$).

• Maximum Kinetic Energy of Alpha (at $\theta = 0^{\circ}$):

  • Using the quadratic relation derived from momentum and energy balance, two values for $t_3$ are obtained:

  • $t_{3, \text{max}} = 1.709\,MeV$ (Alpha goes in direction of neutron).

  • $t_{3, \text{min}} = 0.131\,MeV$.

• Kinetic Energy at $\theta = 90^{\circ}$:

  • Using the simplified formula: $t_3 = \frac{(11 - 1) \times 2 + 11 \times (-0.163)}{4 + 11}$

  • Result: $t_3 = 1.214\,MeV$.

Classification of Radiation

• Non-Ionizing Radiation:

  • Includes visible light, microwaves, infrared (IR), and low-energy UV.

  • These generally do not cause ionization, though high-energy UV can be an exception.

• Ionizing Radiation:

  • Directly Ionizing: Caused by charged particles such as electrons ($\beta^-$), positrons ($\beta^+$), protons ($p$ or $H$), alpha particles (\alpha), and heavy ions (A > 4).

  • Indirectly Ionizing: Includes electromagnetic waves (X-rays, Gamma rays) and neutrons. These particles transfer energy to the medium, which then causes ionization.

Charged Particle Interaction Mechanisms

• Interaction with Matter: Particles primarily interact with the electrons of the target atoms rather than the nucleus.

• Probability Comparison:

  • The area of an atom is defined by the electron orbital size ($\sim 10^{-10}\,m$).

  • The area of the nucleus is significantly smaller ($\sim 10^{-14}\,m$).

  • The probability ratio of interacting with an electron versus the nucleus is roughly $10^4$ or higher ($10^8$ in terms of cross-sectional area); thus, nuclear interactions are often neglected in basic penetration studies.

• Primary Mechanisms:

  1. Ionization: Knocking electrons out of atoms.

  2. Excitation: Raising electrons to higher energy states.

  3. Bremsstrahlung: Emission of electromagnetic radiation due to deceleration.

  4. Cherenkov Radiation: Emission when a particle exceeds the speed of light in a medium.

Cherenkov Radiation

• Mechanism: Occurs when a charged particle moves faster than the speed of light in a specific medium (v > \frac{c}{n}).

• Observation: Often seen as a blue glow in nuclear reactors or spent fuel pools.

• Threshold Energy: For Cherenkov radiation to occur, the particle velocity ratio $\beta$ must be greater than $\frac{1}{n}$, where $n$ is the refractive index.

• Kinetic Energy Relation:

  • $T = (\frac{1}{\sqrt{1 - \beta^2}} - 1) m_0 c^2$

  • Rest mass energy ($m_0 c^2$) for electron: $0.511\,MeV$.

  • Rest mass energy for proton: $938\,MeV$.

• Requirements: To produce Cherenkov radiation, a proton needs $\sim 500\,MeV$, whereas an electron needs only $\sim 0.3\,MeV$.

Ionization and Excitation Processes

• Ionization:

  • Occurs if the incident particle energy is much higher than the ionization potential ($I_p$).

  • Delta Rays: High-energy electrons knocked out of the atom that are capable of causing further secondary ionization.

  • Ion Formation: The atom becomes a heavy positive ion. These ions move slowly and eventually capture electrons to return to a neutral state.

• Excitation:

  • The electron acquires energy to move to a higher orbital (e.g., $E_1 \rightarrow E_2$) if a vacancy exists.

  • The atom stays in an excited state for a very short duration ($10^{-8}$ to $10^{-10}\,s$).

  • Characteristic X-rays: Emitted when the electron drops back to its ground state.

Bremsstrahlung (Braking Radiation)

• Definition: Electromagnetic radiation emitted when a charged particle accelerates or decelerates (often due to direction changes).

• Intensity ($I$): Proportional to the square of acceleration ($a^2$).

• Proportionality: $I \propto \frac{z^2 Z^2}{m^2}$

  • $z$: Charge of the incident particle.

  • $Z$: Atomic number of the medium.

  • $m$: Mass of the incident particle.

• Practical Application:

  • Lighter particles (like electrons/beta particles) produce much higher Bremsstrahlung intensity than heavier alpha particles.

  • High $Z$ materials (like Lead) produce more Bremsstrahlung when shielding beta particles. Therefore, low $Z$ materials (like Plastic) are preferred for beta shielding to minimize the production of secondary X-rays (Buildup factors).

Stopping Power (Bethe-Bloch Relation)

• Definition: The rate of energy loss per unit distance as radiation moves through a medium, given as $\frac{dE}{dx}$.

• Units: $eV/m$ or $MeV/m$.

• The Bethe-Bloch Equation (for heavy particles like protons, alpha):

  • $- \frac{dE}{dx} = \frac{4 \pi r_0^2 m_e c^2 z^2 N Z}{\beta^2} [\ln(\frac{2 m_e c^2 \beta^2 \gamma^2}{I}) - \beta^2]$

• Variables Defined:

  • $r_0$: Classical electron radius ($2.818 \times 10^{-15}\,m$).

  • $4 \pi r_0^2$: $9.98 \times 10^{-29}\,m^2$.

  • $m_e c^2$: Rest energy of electron ($0.511\,MeV$).

  • $\gamma = \frac{T + m_0 c^2}{m_0 c^2} = \frac{1}{\sqrt{1 - \beta^2}}$.

  • $N$: Number density of atoms in medium ($N = \frac{\rho N_A}{A_w}$).

  • $I$: Mean excitation/ionization potential ($I \approx (9.76 + 58.8 Z^{-1.19}) Z$).

• Trends:

  • $\frac{dE}{dx}$ is independent of the mass of the incident particle.

  • Stopping power increases as the speed of the particle decreases (until very low energies).

Numerical Example: Alpha Particle in Silicon

• Parameters:

  • Alpha Energy ($T$): $5\,MeV$.

  • Medium: Silicon ($A_w = 28$, $Z = 14$, $\rho = 2.33 \times 10^3\,kg/m^3$).

• Step 1: Calculate $\gamma$ and $\beta^2$:

  • $m_0 c^2$ for alpha ($4\,amu \times 931.5\,MeV/amu$) $\approx 3726\,MeV$.

  • $\gamma = \frac{5 + 3726}{3726} = 1.00134$.

  • $\beta^2 = 1 - \frac{1}{\gamma^2} = 0.00268$.

• Step 2: Substitution into Stopping Power Formula:

  • Using the constant $4 \pi r_0^2 = 9.98 \times 10^{-29}\,m^2$, charge $z = 2$, and density data.

  • Ionization potential $I$ for Silicon $\approx 172 \times 10^{-6}\,MeV$.

• Result:

  • $\frac{dE}{dx} = 1.48 \times 10^5\,MeV/m$.

  • This value represents the energy deposition rate, which is critical for determining detector thickness or shielding requirements.