Lecture - 5: Radiation Pumping Power, Bremsstrahlung, and Particle Range, and Particle Interactions

Bremsstrahlung Radiation and Intensity

Bremsstrahlung radiation, commonly referred to as "braking radiation," is produced by the acceleration or deceleration of charged particles. This phenomenon occurs when a particle changes its direction or velocity transfer rate, resulting in the emission of electromagnetic radiation.

• Intensity Proportionality: The intensity of Bremsstrahlung radiation is proportional to the square of the particle's acceleration ($a^2$).

• Functional Dependencies: The intensity is a function of the properties of the target material and the properties of the incident particle, specifically its mass.

Stopping Power of Bremsstrahlung

The total stopping power, denoted as $\frac{dE}{dx}$, describes how radiation is stopped by a material. For Bremsstrahlung, this is a complex function. To simplify calculations, it is often expressed as a fraction of the stopping power related to the ionization of that same particle.

• The Ratio Formula: The stopping power due to Bremsstrahlung ($rad$) as a fraction of the stopping power due to ionization ($ion$) is given by the empirical relation:

$(dE/dx)_{rad} =f \times dE/dx_{ion}$

$(dE/dx)_{rad}=\frac{ZT}{750}\times(dE/dx)_{ion}$

Where:

• $Z$ is the atomic number of the medium.

• $T$ is the kinetic energy of the particle in $MeV$.

• Effect of Mass: Bremsstrahlung is more effective with particles of lower mass. Consequently, it has a more significant effect on beta particles, electrons, and positrons than on heavier particles.

• Total Stopping Power: For a beta particle traveling through a material, the total stopping power is the sum of the energy loss due to ionization and the energy loss due to Bremsstrahlung emission:

$\left(\frac{dE}{dx}\right)_{total}=\left(\frac{dE}{dx}\right)_{ion}+\left(\frac{dE}{dx}\right)_{rad}$

This can be rewritten using the ratio formula as:

$\left(\frac{dE}{dx}\right)_{total}=\left(\frac{dE}{dx}\right)_{ion}\times\left[1+\frac{ZT}{750}\right]$

• Initial Kinetic Energy ($T$): For these calculations, the initial kinetic energy of the particle is always used because energy is lost continuously as the particle travels through the material.

• Total Energy Radiated: The total energy radiated as Bremsstrahlung emission is given by the empirical formula:

$T_{rad} = 4.0 \times 10^{-4} \times Z \times T^2$

Numerical Example: Energy Loss in Aluminum vs. Lead

Problem: Consider an electron with an initial kinetic energy ($T$) of $5\,MeV$. What fraction of its energy is lost to Bremsstrahlung radiation as it passes through

(a) Aluminum and

(b) Lead?

(C) What is the Bremsstrahlung energy radiated.

Part A: Aluminum ($Z = 13$)

• $T = 5\,MeV$

• Calculation: $\frac{13 \times 5}{750} = 0.0867$

• Result: The fraction is approximately $0.09$ (or $9\%$) of the ionization radiation.

Part B: Lead ($Z = 82$)

• $T = 5\,MeV$

• Calculation: $\frac{82 \times 5}{750} = 0.5467$

• Result: The fraction is approximately $0.55$ (or $55\%$) of the ionization radiation.

Part C:

$T_{rad} = 4.0 \times 10^{-4} \times 13 \times 5^2$

$=0.13 MeV$

$T_{rad} = 4.0 \times 10^{-4} \times 82 \times 5^2$

$=0.820 MeV$

Conclusion: Bremsstrahlung effect is significantly more predominant in heavy metals ($Z=82$) than in light metals ($Z=13$).

Stopping Power in Compounds and Alloys [13:50]

Most shielding or building materials are not pure elements but are alloys or composite materials. The stopping power for a compound is calculated using the weight percentage of each constituent element.

• Formula for Compound Stopping Power:

$\frac{1}{\rho_{comp}} \times \left(\frac{dE}{dx}\right)_{comp} = \sum w_i \times \frac{1}{\rho_i} \times \left(\frac{dE}{dx}\right)_i$

Where:

• $w_i$ is the weight percentage of element $i$.

• $\rho_i$ is the density of element $i$.

• $\left(\frac{dE}{dx}\right)_i$ is the stopping power of the ionization radiation for element $i$.

Numerical Example: Stopping Power in Air

Problem: Calculate the stopping power of a $10\,MeV$ electron moving through air. Assume air consists of $21\%$ Oxygen and $79\%$ Nitrogen.

For Oxygen ($O$_2, $Z=8$, Atomic Weight = $16$):

Rest Energy for electron:

$m_ec^2=0.511 MeV$

To calculate gamma ($\text{γ}$) for electron with a kinetic energy of $T = 10 \text{ MeV}$, use the formula:

$\text{γ} = \frac{T + m_ec^{2}}{m_ec^{2}}$

$\text{γ} = \frac{10 \text{ MeV} + 0.511 \text{ MeV}}{0.511 \text{ MeV}}$

$= 20.569$

Number density of oxygen (without $\rho$ since that is taken common):

$N=\frac{6.022×10^{23}mol^{−1}}{​{0.016kg/mol}}$
$=3.76\times10^{25}$

Calculate beta:

$\beta² = 1 - \frac{1}{(20.569)^2}$

$β²≈0.99764$

$I = 97.8 \times 10^{-6}$

Use the formula for stopping power of electrons:

$\frac{dE}{dx}=\frac{4 \pi r_0^2 m_e c^2 z^2 N Z}{\beta^2}\left\lbrace\ln(\frac{\beta^{}\gamma\sqrt{^{}\gamma-1}^{}}{I}m_{e}c^2)+\frac{1}{2\gamma^2}\left\lbrack\frac{\left(\gamma-1\right)^2}{Z}+1-\left(\gamma^2+2\gamma-1\right)\ln2\right\rbrack\right\rbrace$
$\frac{dE}{dx}=\frac{4\pi r_0^2(0.511\text{ MeV})(1^2)(3.76\times10^{24})(1)}{0.99764}\left\lbrace\ln\left(\frac{0.9988\cdot20.569\sqrt{20.569-1}}{97.8\times10^{-6}}(0.511\text{ MeV})\right)+\frac{1}{2(20.569)^2}\left[\frac{(20.569-1)^2}{1}+1-(20.569^2+2\cdot20.569-1)\ln2\right]\right\rbrace$

• Calculated Value: $\frac{1}{\rho}\frac{dE}{dx}=0.194\,MeV\,kg^{-1}\,m^2$

  (or 0.196 from code).

For Nitrogen ($N$_2, $Z=7$, Atomic Weight = $14$):

• Calculated Value: $\frac{1}{\rho}\frac{dE}{dx}=0.196\,MeV\,kg^{-1}\,m^2$

Combined for Air:

• Calculation: $(0.21 \times 0.194) + (0.79 \times 0.196)$

  $\rho = 1.29 kg/m³$

$(dE/dx)_{air} = \rho \times [(0.21 \times 0.194) + (0.79 \times 0.196)]$

• Final Result provided in transcript: $0.253\,MeV$

Concept of Range and Idealized Experiment [29:00]

The "range" describes the distance a particle travels in a material before losing all its energy.

• Normalized Range: Often expressed in units of $kg/m^2$ to normalize for the density of the material.

• Linear Range: The actual distance traveled in meters ($m$). It is found by dividing the normalized range by the material density ($\rho$).

• Idealized Experiment: In a scenario where all particles travel in the same direction (transverse) through a material of thickness $x$, the number of particles passing through ($n$) is measured against the initial count ($n_0$).

• Range Definition ($R_{50}$): The thickness required to decrease the number of particles by $50\%$.

Range of Alpha Particles in Air [32:20]

The range of alpha particles in air can be calculated using different empirical relations based on their initial kinetic energy ($T$ in $MeV$):

1. Low Energy Range ($1$ to $4\,MeV$):     $R_{air}\,(mm) = e^{1.61\sqrt{T}}$

2. Higher Energy Range:     $R = (0.05T + 2.85)T^{3/2}$

The Bragg-Kleeman Rule for Range Conversion [34:42]

If the range of a particle is known in one medium (e.g., air), it can be converted to calculate the range in another medium using the Bragg-Kleeman rule:

$\frac{R_1 \rho_1}{\sqrt{A_1}} = \frac{R_2 \rho_2}{\sqrt{A_2}}$

Where:

• $R$ is the range.

• $\rho$ is the density.

• $A$ is the atomic mass number.

This rule is applicable to alpha particles, protons, and other heavy particles, but cannot be used for electrons or positrons.

Effective Atomic Mass Number ($A_{eff}$) [38:00]

For compounds, the effective square root of the atomic mass is required for the Bragg-Kleeman rule:

$\sqrt{A_{eff}} = \frac{1}{\sum^L_{i=1} (w_i / \sqrt{A_i})}$

• Water ($H_2O$): The effective square root of atomic mass is approximately $3$, making $A_{eff} \approx 9$.

• Air: The effective atomic mass number ($A_{eff}$) is approximately $14.7$, and $\sqrt{A_{eff}} = 3.84$.

Numerical Example: Alpha Range in Gold [45:00]

Problem: Determine the range of a $3\,MeV$ alpha particle in gold ($Z = 79$, $A = 197$, $\rho = 19.32 \times 10^3\,kg/m^3$).

1. Range in Air:     $R_{air} = e^{1.61\sqrt{3}} \approx 16.25\,mm$

2. Conversion to Gold:     

$\frac{R_{1(mm)} \rho_1}{\sqrt{A_1}} = \frac{R_{air} \rho_2}{\sqrt{A_2}}$

$R_{1(mm)} = 16.25\cdot \left( \frac{1.29}{19.32 \times 10^{3}} \right)\cdot \sqrt{\frac{197}{14.73}}$

Using experimental data and the Bragg-Kleeman rule, the calculated empirical range is approximately $3.8\,\mu m$. Experimental results for alpha particles in silicon ($R=12.5\,\mu m$) converted to gold yield approximately $4\,\mu m$. Both values are very close.

Range of Protons (Bischel's Relation) [51:41]

Proton range in materials like aluminum is calculated using Bischel's Relation:

1. Energy between $1\,MeV$ and $2.7\,MeV$:     $R\,(\mu m) = 14.21 \times T^{1.587}$

2. Energy between $2.7\,MeV$ and $20\,MeV$:     $R = 10.5 \times \frac{T^{1.5}}{0.68 + 0.434\ln(T)}$

3. Instructor mentioned to use Bragg-Kleeman for range conversion.

The Same-Speed Formula for Same-Speed Particles

This formula allows for calculating the range of protons or neutrons/deuterons based on the range of alpha particles provided they are moving at the same speed:

$R_x = 4 \times \frac{M_x}{M_{\alpha}} \times (R_{\alpha} - 2\,mm)$

Where:

• $M_x$ is the mass of the particle (proton or neutron).

• $M_{\alpha}$ is the mass of the alpha particle.

• $R_{\alpha}$ is the range of an alpha particle at the same speed.

Numerical Example: Deuteron Range in Air [55:56]

Problem: Calculate the range of a $5\,MeV$ deuteron ($H^2_1$) in air using the same-speed requirement.

$T_d=\frac{1}{2}\times M_d \times V_d²$

$T_\alpha=\frac{1}{2}\times M_\alpha \times V_\alpha²$

for same speed $V_\alpha = V_d$ :

$\frac{T_d}{T_\alpha} = \frac{M_d}{M_\alpha} = \frac{2}{4}$

• Identify Velocity Equality:

            For speed $v_D = v_{\alpha}$ , the kinetic energy of the alpha particle must be twice that of             the deuteron because $T = \frac{1}{2}mv^2$ and the mass ratio is approximately $2:4$.             Thus, if $T_D = 5\,MeV$, use $T_{\alpha} = 10\,MeV$.

• Find Alpha Range at $10\,MeV$:

            Under these conditions, $R_{\alpha} = 106\,mm$.

• Calculate Deuteron Range:

    $R_D = 4 \times \frac{2}{4} \times (106 - 2) = 208\,mm$

            Now Bragg Kleeman can be used to find the range in other materials.

Range of Electrons and Positrons (Tabata Formula) [1:03:15]

Lighter particles like electrons and positrons do not exhibit a range plateau; their transmission intensity decreases continuously until reaching background levels.

• Tabata Formula: Valid for energies between $0.3\,keV$ and $30\,MeV$. It involves a complex empirical relation using five constants ($a_1$ through $a_5$) and the relativistic factor $\gamma$.

$R(\mathrm{kg}/\mathrm{m}^2) = a_1\left(\frac{\ln(1+a_2(\gamma-1))}{a_2} - \frac{a_3(\gamma-1)}{1+a_4(\gamma-1)^{a_5}}\right)$

$a_1 = \frac{2.335A}{Z^{1.209}}$

$a_2 = 1.78 \times 10^{-4} Z$

$a_3 = 0.9891 - (3.01 \times 10^{-4} Z)$

$a_4 = 1.468 - (1.180 \times 10^{-2} Z)$

$a_5 = \frac{1.232}{Z^{0.109}}$

• Energy Representation: Uses $\gamma$ as the ratio between total energy and rest mass.

• Effective Compound Units: Similar to other models, this can be extended to compound materials by identifying a $Z_{effective}$.

Range and Attenuation of Beta Particles [1:10:44]

Beta particles transmit through materials in a manner similar to electrons but follow an exponential decrease in intensity, defined by the attenuation coefficient ($\mu$).

• Transmission Formula:    

 $N_t = N_0 \times e^{-\mu t}$

• Attenuation Coefficient ($\mu$):     $\mu = 1.7 \times E_{max}^{-1.14}$     Units: $m^2/kg$. Energy must be in $MeV$.

Numerical Example: Beta Particle Transmission in Aluminum [1:13:50]

Problem: What fraction of $2\,MeV$ beta particles pass through an aluminum foil of $0.1\,mm$ thickness?

• Calculate Mu ($\mu$):  

   $\mu = 1.7 \times (2)^{-1.14} \approx 0.7714\,m^2/kg$

• Calculating the Exponent ($\mu t$):     

Integrating density ($\rho_{Al} = 2.7 \times 10^3\,kg/m^3$) and thickness ($t = 0.1 \times 10^{-3}\,m$):     $\mu \times \rho \times t = 0.7714 \times 2.7 \times 0.1 \approx 0.208$

• Final Fraction ($e^{-0.208}$):     Result: Approximately $81\%$ of the particles will pass through the foil.