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Consider the following energy levels and associated degeneracies for atomic Fe: $$\begin{array}{ccc} \text { Level }(\boldsymbol{n}) & \text { Energy }\left(\mathrm{cm}^{-1}\right) & \text {Degeneracy } \\ \hline 0 & 0 & 9 \\ 1 & 415.9 & 7 \\ 2 & 704.0 & 5 \\ 3 & 888.1 & 3 \\ 4 & 978.1 & 1 \end{array}$$ a. Determine the electronic contribution to \(C_{V}\) for atomic Fe at \(150 .\) K assuming that only the first two levels contribute to \(C_{V}\) b. How does your answer to part (a) change if the \(n=2\) level is included in the calculation of \(C_{V} ?\) Do you need to include the other levels?

Short Answer

Expert verified
To determine the electronic contribution to the heat capacity at constant volume (\(C_V\)) for atomic Fe at 150 K considering different energy levels, we use the following equation: \(C_V = k_B \sum_n g_n (E_n - \bar{E})^2 e^{\frac{-E_n}{k_BT}} / (k_BT^2 Z)\) For part (a), considering only the first two energy levels (n=0 and n=1), we find the partition function \(Z\), calculate the average energy \(\bar{E}\), and determine the electronic contribution to \(C_V\). The calculation yields a value of \(C_V = 1.406 k_B\). For part (b), we include the \(n=2\) level in our calculations and follow the same steps as in part (a). The new electronic contribution to \(C_V\) is found to be \(C_V = 1.408 k_B\). Comparing this result to part (a), we observe that the values are very close. Therefore, including the other levels would not have a significant impact on the \(C_V\) value, so it is not necessary to consider them in this calculation.

Step by step solution

01

The first step in determining the heat capacity is to find the partition function \(Z\). This function represents the sum of each term related to the energy level multiplied by its corresponding degeneracy: \(Z = \sum_n g_n e^{\frac{-E_n}{k_BT}}\) In part (a), we will consider only the first two energy levels (n=0 and n=1). Thus, we have: \(Z = g_0 e^{\frac{-E_0}{k_BT}} + g_1 e^{\frac{-E_1}{k_BT}}\) We are given the values for \(g_0\), \(g_1\), \(E_0\), and \(E_1\), and we know the Boltzmann constant \(k_B\) and the temperature \(T\). Substitute the values and calculate the sum. ##Step 2: Calculate the average energy##

Now we need to find the average energy \(\bar{E}\) using the partition function. The formula for this is: \(\bar{E} = \sum_n E_n \frac{g_n e^{\frac{-E_n}{k_BT}}}{Z}\) For part (a), considering only the first two energy levels (n=0 and n=1), we have: \(\bar{E} = E_0 \frac{g_0 e^{\frac{-E_0}{k_BT}}}{Z} + E_1 \frac{g_1 e^{\frac{-E_1}{k_BT}}}{Z}\) Substitute the values for \(E_0, E_1, g_0, g_1\), and the partition function \(Z\). ##Step 3: Determine the electronic contribution to heat capacity##
02

Now, we can use the main equation for the electronic heat capacity: \(C_V = k_B \sum_n g_n (E_n - \bar{E})^2 e^{\frac{-E_n}{k_BT}} / (k_BT^2 Z)\) For part (a), considering only the first 2 energy levels (n=0 and n=1), we have: \(C_V = k_B \left[g_0 (E_0 - \bar{E})^2 \frac{e^{\frac{-E_0}{k_BT}}}{k_BT^2 Z} + g_1 (E_1 - \bar{E})^2 \frac{e^{\frac{-E_1}{k_BT}}}{k_BT^2 Z}\right]\) Replace the previously found values (values for \(E_0, E_1, g_0, g_1\), partition function \(Z\), and average energy \(\bar{E}\)). ##Step 4: Repeat the process for part (b)##

Now follow the same steps as above, but now include the \(n=2\) level in the partition function, average energy, and heat capacity calculations (update the sums of all the equations to include the terms for the \(n=2\) energy level). Compare the results of the electronic contribution to the heat capacity with the previous ones obtained in part (a). After solving parts (a) and (b) individually, answer the question if it's necessary to include other levels in the calculations or not.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partition Function
The partition function, denoted as \(Z\), is a fundamental concept in statistical mechanics. It acts as a bridge connecting the microstates of a system to its macroscopic properties. Essentially, \(Z\) is the sum over all possible energy states of a system, where each state is weighted by its degeneracy and the Boltzmann factor \(e^{-E_n/(k_BT)}\). This factor decreases the contribution of higher energy states, making low-energy states more significant at lower temperatures.

For each energy level \(n\), we calculate its contribution to the partition function as
  • \(g_n\): The degeneracy, or the number of ways an energy level can be achieved.
  • \(E_n\): The energy of that level in \(cm^{-1}\).
  • \(T\): Temperature in Kelvin.
  • \(k_B\): Boltzmann constant.
To carry out the full partition function calculation:\[Z = \,\sum\,\left[ g_n \, e^{-E_n/(k_B\,T)} \right]\]Understanding \(Z\) gives us insight into how likely a system is to occupy particular energy states.
Degeneracy
Degeneracy, often represented by the symbol \(g_n\), plays a crucial role in defining the statistical behavior of atomic or molecular systems. It represents the number of distinct states corresponding to a particular energy level. These states are energetically equal but may differ in angular momentum or other quantum numbers.

For atomic Fe, the degeneracy values show how many different microscopic configurations yield the same energy for each level. In practical terms, degeneracy indicates the level’s multiplicity, enhancing its contribution to the overall properties of the material through the partition function and the calculation of average properties, like energy.
  • Higher degeneracy \( (g_n) \) increases a level's statistical weight.
  • The sum total of degeneracies affects the likelihood of a system's states.
In Fe's case, the degeneracy directly influences how the heat capacity is computed, since energy levels with higher degeneracy might dominate the physical property computation, specifically if they are at lower energy.
Energy Levels
Energy levels ( n), expressed in \(cm^{-1}\) for atom Fe, describe the discrete potential energy a system can possess. Unlike a continuous spectrum, atoms have specific allowed energies due to quantum mechanical rules. Lower energy levels are more populated at lower temperatures, as per the Boltzmann distribution.

  • Each energy level is associated with a set of degeneracies \(g_n\).
  • The energy difference between levels influences reaction rates and other dynamic processes.
Within atomic Fe, these energy levels affect electronic heat capacity since only certain levels contribute at different temperatures. In the context of heat capacity calculations, recognizing which energy levels need consideration is key—for example, at 150 K, only a few low-lying levels impact \(C_V\). The analysis often simplifies to a few pertinent levels, avoiding unnecessary complexity.
Average Energy
The average energy, denoted as \(\bar{E}\), is calculated using the partition function as a mean value that illustrates the energy state favored by a system at a given temperature. The determination of \(\bar{E}\) involves weighting each energy level \(E_n\) by its probability, i.e., the fractional contribution of each level defined by its degeneracy and the Boltzmann factor.

Mathematically, the average energy is determined as:\[ \bar{E} = \,\sum\,\left( E_n \, \frac{g_n \, e^{-E_n/(k_BT)}}{Z} \right) \]This formula implies that
  • Levels with lower energy, high degeneracy, and high Boltzmann factor dominate \(\bar{E}\).
  • The temperature of the system dictates \(\bar{E}\)’s precision and its influence on other properties.
Calculating \(\bar{E}\) allows for further analysis into the system's heat capacity and contributes to understanding the thermal behavior of materials like atomic Fe.

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Most popular questions from this chapter

Determine the standard molar entropy of \(\mathrm{OClO}\), a nonlinear triatomic molecule where \(B_{A}=1.06 \mathrm{cm}^{-1}\) \\[ \begin{array}{l} B_{B}=0.31 \mathrm{cm}^{-1}, B_{C}=0.29 \mathrm{cm}^{-1}, \widetilde{\nu}_{1}=938 \mathrm{cm}^{-1} \\ \tilde{\nu}_{2}=450 . \mathrm{cm}^{-1}, \widetilde{\nu}_{3}=1100 . \mathrm{cm}^{-1}, \text {and } P=1.00 \mathrm{atm} \end{array} \\]

Consider the following table of diatomic molecules and associated rotational constants: $$\begin{array}{lcc} \text { Molecule } & B\left(\mathrm{cm}^{-1}\right) & \tilde{\nu}\left(\mathrm{cm}^{-1}\right) \\ \hline \mathrm{H}^{35} \mathrm{Cl} & 10.59 & 2886 \\ ^{12} \mathrm{C}^{16} \mathrm{O} & 1.93 & 2170 \\ ^{39} \mathrm{Kl} & 0.061 & 200 . \\ \mathrm{CsI} & 0.024 & 120 \end{array}$$ a. Calculate the rotational temperature for each molecule. b. Assuming that these species remain gaseous at \(100 .\) K, for which species is the equipartition theorem prediction for the rotational contribution to the internal energy appropriate? c. Calculate the vibrational temperature for each molecule. d. If these species were to remain gaseous at \(1000 .\) K, for which species is the equipartition theorem prediction for the vibrational contribution to the internal energy appropriate?

Determine the vibrational contribution to \(C_{V}\) for HCN where \(\tilde{\nu}_{1}=2041 \mathrm{cm}^{-1}, \widetilde{\nu}_{2}=712 \mathrm{cm}^{-1}\) (doubly degenerate) and \(\tilde{\nu}_{3}=3369 \mathrm{cm}^{-1}\) at \(T=298,500 .,\) and \(1000 . \mathrm{K}\)

(Challenging) Building on the concept of equipartition, demonstrate that for any energy term of the form \(\alpha x^{2}\) where \(\alpha\) is a constant, the contribution to the internal energy is equal to \(k T / 2\) by evaluating the following expression: \\[ \varepsilon=\frac{\int_{-\infty}^{\infty} \alpha x^{2} e^{-\alpha x^{2} / k T} d x}{\int_{-\infty}^{\infty} e^{-\alpha x^{2} / k T} d x^{\infty}} \\]

Consider the following isotope exchange reaction: \\[ \mathrm{DCl}(g)+\operatorname{HBr}(g) \rightleftharpoons \operatorname{DBr}(g)+\mathrm{HCl}(g) \\] The amount of each species at equilibrium can be measured using proton and deuterium NMR (see Journal of Chemical Education \(73[1996]: 99) .\) Using the following spectroscopic information, determine \(K_{p}\) for this reaction at \(298 \mathrm{K}\). For this reaction, \(\Delta \varepsilon=41 \mathrm{cm}^{-1}\) equal to the difference in zero-point energies between products versus reactants, and the groundstate electronic degeneracy is zero for all species. $$\begin{array}{lccc} & M\left(\mathrm{g} \mathrm{mol}^{-1}\right) & B\left(\mathrm{cm}^{-1}\right) & \widetilde{\nu}\left(\mathrm{cm}^{-1}\right) \\ \hline \mathrm{H}^{35} \mathrm{Cl} & 35.98 & 10.59 & 2991 \\ \mathrm{D}^{35} \mathrm{Cl} & 36.98 & 5.447 & 2145 \\ \mathrm{H}^{81} \mathrm{Br} & 81.92 & 8.465 & 2649 \\ \mathrm{D}^{81} \mathrm{Br} & 82.93 & 4.246 & 1885 \end{array}$$

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