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]