# Difference between revisions of "Boltzmann distribution"

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\begin{document} | \begin{document} | ||

\title{Boltzmann Distribution} | \title{Boltzmann Distribution} | ||

+ | |||

+ | \section{Definition} | ||

+ | The Boltzmann distribution gives the relative fraction of atoms in two states in thermal equilibrium at a certain temperature, taking into account the degeneracies of these states and the energy difference between states. This distribution applies to large ensembles of atoms such that statistical arguments are valid. The formula for the Boltzmann distribution is given by the following: | ||

+ | |||

+ | $$\frac{n_2}{n_1}=\frac{g_2}{g_1}e^{-\frac{\Delta E}{kT}}$$ | ||

+ | |||

+ | where $n_1,n_2$ are the number (or number density) of atoms in energy states 1 and 2, respectively, $g_1,g_2$ are the respective degeneracies of those energy states (i.e., how many distinct configurations of the atom have that same energy state), $\Delta E=E_2-E_1$ is the energy difference between the two states, $k$ is Boltzmann's constant, and $T$ is the temperature describing the distribution of states in the system | ||

+ | |||

+ | \section{Derivation} | ||

+ | |||

+ | Consider an isolated statistical ensemble of N particles with energies given by | ||

+ | |||

+ | $$E = \sum \epsilon_i n_i$$ | ||

+ | |||

+ | and total number of particles | ||

+ | |||

+ | $$N = \sum n_i.$$ | ||

+ | |||

+ | Then, the N particles will be arranged into microstates given by | ||

+ | |||

+ | $$W = (g_1^{n_1} g_2^{n_2} g_3^{n_3} ...)\frac{N!}{n_1!n_2!n_3! ...}.$$ | ||

+ | |||

+ | Solving for lnW, we then get | ||

+ | |||

+ | $$lnW = n_1lng_1 + n_2lng_2 + ... + lnN! - n_1lnn_1 - n_2lnn_2 - ... + n_1 + n_2 + ...$$ | ||

+ | $$lnW = n_1(1 + ln\frac{g_1}{n_1}) + n_2(1 + ln\frac{g_2}{n_2}) + ... + lnN!$$ | ||

+ | |||

+ | where we implemented Stirling's approximation for large N, $lnN! \approx NlnN - N$, to simplify our expression. Using Lagrange's method of undetermined multipliers, we seek a maximum in | ||

+ | |||

+ | $$ lnW - \alpha \sum n_i - \beta\sum\epsilon_i\n_i $$ | ||

+ | |||

+ | |||

+ | \section{Degeneracies of the hydrogen atom} | ||

+ | |||

+ | The energy degeneracies corresponding to each E_n for the hydrogen atom are given by: | ||

+ | |||

+ | \begin{equation} | ||

+ | g_n = \sum^{n-1}_{l=0} (2l + 1) = n^2 | ||

+ | \end{equation} | ||

+ | |||

+ | Taking the two electron spin states of hydrogen into account, our degeneracies then ultimately become | ||

\begin{equation} | \begin{equation} | ||

− | + | g_n = 2n^2. | |

\end{equation} | \end{equation} | ||

− | |||

\end{document} | \end{document} | ||

</latex> | </latex> |

## Revision as of 15:55, 23 November 2016

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# Boltzmann Distribution

## 1 Definition

The Boltzmann distribution gives the relative fraction of atoms in two states in thermal equilibrium at a certain temperature, taking into account the degeneracies of these states and the energy difference between states. This distribution applies to large ensembles of atoms such that statistical arguments are valid. The formula for the Boltzmann distribution is given by the following:

where are the number (or number density) of atoms in energy states 1 and 2, respectively, are the respective degeneracies of those energy states (i.e., how many distinct configurations of the atom have that same energy state), is the energy difference between the two states, is Boltzmann’s constant, and is the temperature describing the distribution of states in the system

## 2 Derivation

Consider an isolated statistical ensemble of N particles with energies given by

and total number of particles

Then, the N particles will be arranged into microstates given by

Solving for lnW, we then get

where we implemented Stirling’s approximation for large N, , to simplify our expression. Using Lagrange’s method of undetermined multipliers, we seek a maximum in

**Failed to parse (unknown function "\n"): {\displaystyle lnW - \alpha \sum n_ i - \beta \sum \epsilon _ i\n _ i \,\!}**

## 3 Degeneracies of the hydrogen atom

The energy degeneracies corresponding to each E_n for the hydrogen atom are given by:

Taking the two electron spin states of hydrogen into account, our degeneracies then ultimately become