Statistical Thermodynamics and Rate Theories/Translational energy

A gas particle can move through space in three independent directions: x, y, and z. Each of these directions is a distinct translational degree of freedom. For ideal gases, these particles can move freely through space in any of these directions until it collides with the wall of its container. For isolated systems, collisions with the wall of the container are assumed to be elastic, meaning no energy is lost upon impact.

Particle in a 1D Box
The first quantum mechanical model used for describing translation is a particle in a simple 1D box. It is free to move anywhere along one axis (usually assigned to be the x axis) between the arbitrarily assigned boundary limits 0 and a. At distances smaller than 0 and greater than “a” the potential energy function immediately rises to infinity. The particle cannot move past these points. 0 and a represent the walls of the 1D box. We can mathematically assign boundary conditions using this information which allows the derivation of the wave function for a particle in a 1D box. The resulting piece-wise function, wave function and energy level equation are given as follows,


 * $$\mathcal{V}(x) = \begin{cases}

\infty, x<0 \\ 0, 0{\leq}x{\leq}a \\ \infty, x>a \end{cases} $$


 * $$\Psi(x) = \sqrt{\frac{2}{a}} \sin{\left( \frac{n\pi x}{a} \right)}$$
 * $$n = 1,2,3,...$$

As well as the corresponding energy levels of the system.


 * $$\epsilon_n = \frac{h^2 n^2}{8ma^2}$$

Note: In the equations above m represents the mass of the gas particle, and h is Planck's constant.

Zero-Point Energy
The energy levels for translation within the box are quantized, only discrete energy levels can be occupied by the particle at any given time. Each of these energy levels are defined by a single quantum number n. n can take on any integer values starting from 1 and up to a hypothetical infinity. The n = 0 state for a particle in a box does not exist. The particle, in accordance with the Heisenberg uncertainty principle cannot be motionless. If this were to be the case then both the momentum and position of the particle could be determined simultaneously, which is a violation of the principle. The energy of a particle in a 1D box at the lowest translational energy level is therefore non-zero (i.e., n = 1).

Probability Density Plots and the Correspondence Principle
It is possible to construct a probability density plot of the particle in a 1D box wave function. The plot is characterized by sizable “humps” at low values of n. For example, at n=1 there is one large hump which spans from a minimum at x = 0, a maximum somewhere in the center and a second minimum at x = a. These minimums represent regions of zero probability density. In other words, the particle will not be found in these regions.

As n increases the spacing between humps of the probability density plot becomes smaller and smaller until a near continuum is achieved at sufficiently high values of n. It must also be noted that the magnitude of the energy spacing’s between translational energy levels is extremely small relative to the amount of energy available to a particle under normal conditions. At room temperature ideal gas particles will occupy very high energy levels. As we increase the quantum number to a large enough value the behavior of the system will begin to reproduce that of classical mechanics in that the particle is essentially equally likely to be found anywhere in the box rather than within discrete hump regions seen at lower values of n. This is what is known as the correspondence principle.

Particle in a 3D Box
The correspondence principle applies to a 3D box as well. For a particle in a 3D box, the particle is able to move in any direction along the x, y and z axes. Since its motion now incorporates a combination of three possible directions the system must include two additional quantum numbers to compensate for the difference. nx, ny, and nz for the x, y, and z dimensions, respectively. Both the wave function and the equation for the energy levels must be adjusted to compensate for the new quantum numbers. These are given as follows,


 * $$\Psi_{n_x,n_y,n_z} = \sqrt{\frac{8}{abc}} \sin \left(\frac{n_x{\pi}x}{a} \right) \sin \left (\frac{n_y{\pi}y}{b} \right) \sin \left(\frac{n_z{\pi}z}{c} \right)$$


 * $$\epsilon_{n_x,n_y,n_z} = \frac{h^2}{8m} \left({\frac{{n_x}^2}{a^2}}+{\frac{{n_y}^2}{b^2}}+{\frac{{n_z}^2}{c^2}}\right)$$

a, b, and c represent the length of each corresponding side of the box. If the box were a cube then the energy level equation can be simplified because sides a, b, and c of the cube are all equal. The resulting equation would take on the form,


 * $$\epsilon_{n_x,n_y,n_z} = \frac{h^2}{8m{a^2}} ({{n_x}^2}+{{n_y}^2}+{{n_z}^2})$$

For a particle in a cube some combinations of quantum numbers will give the same energy level. Energy levels with a different set of quantum numbers but having the same energy are said to be degenerate and unless all three of the quantum numbers are identical there is always another combination of the three quantum numbers that will give rise to a degenerate state.

Example
Calculate the energy difference, $$\Delta E$$, for the translation of N2 from the ground state to the first excited state. Assume the box is a cube with an edge length of 10 cm.

Solution
The ground state, where $$n_x=n_y=n_z=1$$, has a degeneracy of 1. The first excited state has a degeneracy of 3, involving the different combinations of quantum numbers $$n=1,1,$$ and 2. Can be determined by using the following equation:
 * $$\epsilon_{n_x,n_y,n_z} = \frac{h^2}{8m} \left({\frac{{n_x}^2}{a^2}}+{\frac{{n_y}^2}{b^2}}+{\frac{{n_z}^2}{c^2}} \right)$$

However because all edges of the cube is 10 cm, then a=b=c=10 cm = 0.1 m. Then the equation becomes:


 * $$\epsilon_{n_x,n_y,n_z} = \frac{h^2}{8m{a^2}} \left({{n_x}^2}+{{n_y}^2}+{{n_z}^2} \right)$$

The mass of N2 can be determined with the following equation:


 * $$\mu = \cfrac{m_N m_N}{m_N + m_N},\!\,$$
 * $$\mu = \cfrac{m_N^2}{2 m_N},\!\,$$
 * $$\mu = \cfrac{m_N}{2},\!\,$$
 * $$\mu = \cfrac{14.0067 u}{2},\!\,$$
 * $$\mu = (7.00335 u)(1.660549 \times 10^{-27}kg/u),\!\,$$
 * $$\mu = 1.16294 \times 10^{-26} kg,\!\,$$

Using the mass the $$\Delta E$$ can now be determined.


 * $$\epsilon_{n_x,n_y,n_z} = \frac{h^2}{8m{a^2}} ({{n_x}^2}+{{n_y}^2}+{{n_z}^2})-({{n_x}^2}+{{n_y}^2}+{{n_z}^2})$$
 * $$\epsilon_{n_x,n_y,n_z} = \frac{({6.626 \times 10^{-34}})^2 J s}{8(1.16294 \times 10^{-26} kg){0.1 m^2}} ({{2}^2}+{{1}^2}+{{1}^2})-({{1}^2}+{{1}^2}+{{1}^2})$$
 * $$\epsilon_{n_x,n_y,n_z} = \frac{{4.3904 \times 10^{-67}} J^2 s^2}{9.30352 \times 10^{-28}kg m^2} (6-3)$$
 * $$\epsilon_{n_x,n_y,n_z} = 1.4157 \times 10^{-39}J$$