Simulations of Particle Interactions with Matter

Overview

Particle Detection

A device detects a particle only after the particle transfers energy to the device.

Energy intrinsic to a device depends on the material used in a device

Some device of material with an average atomic number ([math]Z[/math]) is at some temperature ([math]T[/math]). The materials atoms are in constant thermal motion (unless T = zero degrees Klevin).

Statistical Thermodynamics tells us that the canonical energy distribution of the atoms is given by the Maxwell-Boltzmann statistics such that

[math]P(E) = \frac{1}{kT} e^{-\frac{E}{kT}}[/math]

[math]P(E)[/math] represents the probability of any atom in the system having an energy [math]E[/math] where

[math]k= 1.38 \times 10^{-23} \frac{J}{mole \cdot K}[/math]

Note: You may be more familiar with the Maxwell-Boltzmann distribution in the form

[math]N(\nu) = 4 \pi N \left ( \frac{m}{2\pi k T} \right ) ^{3/2} v^2 e^{-mv^2/2kT}[/math]

where [math]N(v) \Delta v[/math] would represent the molesules in the gas sample with speeds between [math]v[/math] and [math]v + \Delta v[/math]

Example 1: P(E=5 eV)

What is the probability that an atom in a 12.011 gram block of carbon would have and energy of 5 eV?

First lets check that the probability distribution is Normailized; ie: does [math]\int_0^{\infty} P(E) dE =1[/math]?

[math]\int_0^{\infty} P(E) dE = \int_0^{\infty} \frac{1}{kT} e^{-\frac{E}{kT}} dE = \frac{1}{kT} \frac{1}{\frac{1}{-kT}} e^{-\frac{E}{kT}} \mid_0^{\infty} = - [e^{-\infty} - e^0]= 1[/math]

[math]P(E=5eV)[/math] is calculated by integrating P(E) over some energy interval ( ie:[math] N(v) dv[/math]). I will arbitrarily choose 4.9 eV to 5.1 eV as a starting point.

[math]\int_{4.9 eV}^{5.1 eV} P(E) dE = - [e^{-5.1 eV/kT} - e^{4.9 eV/kT}][/math]

[math]k= (1.38 \times 10^{-23} \frac{J}{mole \cdot K} ) = (1.38 \times 10^{-23} \frac{J}{mole \cdot K} )(6.42 \times 10^{18} \frac{eV}{J})= 8.614 \times 10^{-5} \frac {eV}{mole \cdot K}[/math]

assuming a room empterature of [math]T=300 K[/math]

then[math]kT = 0.0258 \frac{eV}{mole}[/math]

and

[math]\int_{4.9 eV}^{5.1 eV} P(E) dE = - [e^{-5.1/0.0258} - e^{4.9/0.0258}] = 4.48 \times 10^{-83} - 1.9 \times 10^{-86} \approx 4.48 \times 10^{-83}[/math]

or in other words the precise mathematical calculation of the probability may be approximated by just using the distribution function alone

[math]P(E=5eV) = e^{-5/0.0258} \approx 10^{-85}[/math]

This approximation breaks down as [math]E \rightarrow 0.0258 eV[/math]

Since we have 12.011 grams of carbon and 1 mole of carbon = 12.011 g = [math]6 \times 10^{23} [/math]carbon atoms

We do not expect to see a 5 eV carbon atom in a sample size of [math]6 \times 10^{23} [/math] carbon atoms when the probability of observing such an atom is [math]\approx 10^{-85}[/math]

The energy we expect to see would be calculated by

[math]\lt E\gt = \int_{0}^{\infty} E \cdot P(E) dE[/math]

If you used this block of carbon as a detector you would easily notice an event in which a carbon atom absorbed 5 eV of energy as compared to the energy of a typical atom in the carbon block.

---

Silicon detectors and Ionization chambers are two commonly used devices for detecting radiation.

approximately 1 eV of energy is all that you need to create an electron-ion pair in Silicon

[math]P(E=1 eV) = e^{-1/0.0258} \approx 10^{-17}[/math]

approximately 10 eV of energy is needed to ionize an atom in a gas chamber

[math]P(E=10 eV) = e^{-10/0.0258} \approx 10^{-169}[/math]

The low probability of having an atom with 10 eV of energy means that an ionization chamber would have a better Signal to Noise ratio (SNR) for detecting 10 eV radiation than a silicon detector

But if you cool the silicon detector to 200 degrees Kelvin (200 K) then

[math]P(E=1 eV) = e^{-1/0.0172} \approx 10^{-26} 10^{-17}\lt \lt [/math]

So cooling your detector will slow the atoms down making it more noticable when one of the atoms absorbs energy.

also, if the radiation flux is large, more electron-hole pairs are created and you get a more noticeable signal.

Unfortunately, with some detectore, like silicon, you can cause radiation damage that diminishes it quantum efficiency for absorbing energy.

The Monte Carlo method

Stochastic

from the greek word "stachos"

a means of, relating to, or characterized by conjecture and randomness.

A stochastic process is one whose behavior is non-deterministic in that the next state of the process is partially determined.

Physics has many such non-deterministic systems:

• Quantum Mechanics
• Thermodynamics

Basically the monte-carlo method uses a random number generator (RNG) to generate a distribution (gaussian, uniform, Poission,...) which is used to solve a stochastic process based on an astochastic description.

Example 2 Calculation of [math]\pi[/math]

Astochastic description

[math]\pi[/math] may be measured as the ratio of the area of a circle of radius [math]r[/math] divided by the area of a square of length [math]2r[/math]

[math]\frac{A_{circle}}{A_{square}} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4}[/math]