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Consider an ensemble of units in which the first excited electronic state at energy \(\varepsilon_{1}\) is \(m_{1}\) -fold degenerate, and the energy of the ground state is \(m_{o}\) -fold degenerate with energy \(\varepsilon_{0}\) a. Demonstrate that if \(\varepsilon_{0}=0,\) the expression for the electronic partition function is \\[ q_{E}=m_{o}\left(1+\frac{m_{1}}{m_{o}} e^{-\varepsilon_{1} / k T}\right) \\]\ b. Determine the expression for the internal energy \(U\) of an ensemble of \(N\) such units. What is the limiting value of \(U\) as the temperature approaches zero and infinity?

Short Answer

Expert verified
The electronic partition function for the given ensemble of units is: \\[ q_{E} = m_0 \left(1 + \frac{m_1}{m_0} e^{-\varepsilon_1 / k T}\right) \\] The internal energy of the ensemble is: \\[ U = -\left(-\frac{m_1 \varepsilon_1 e^{-\varepsilon_1 \beta}}{m_0 + m_1 e^{-\varepsilon_1 \beta}}\right) \\] As the temperature approaches zero, the internal energy approaches 0. As the temperature approaches infinity, the internal energy approaches \(\frac{m_1}{m_0 + m_1} \varepsilon_1\).

Step by step solution

01

Find the electronic partition function

To find the electronic partition function, we can use the following equation: \\[ q_{E} = \sum_{i} g_i e^{-\varepsilon_i / k T} \\] Here, \(g_i\) is the degeneracy of energy level \(\varepsilon_i\), \(k\) is the Boltzmann constant, and \(T\) is the temperature. Since there are only two energy levels in this problem, the partition function can be written as: \\[ q_{E} = g_0 e^{-\varepsilon_0 / k T} + g_1 e^{-\varepsilon_1 / k T} \\] Given that \(\varepsilon_0 = 0\), the first term simplifies to: \\[ g_0 e^{-\varepsilon_0 / k T} = g_0 e^0 = g_0 = m_0 \\] Substitute this and the given values for \(g_0\) and \(g_1\) into the partition function equation, obtaining: \\[ q_{E} = m_0 \left(1 + \frac{m_1}{m_0} e^{-\varepsilon_1 / k T}\right) \\]
02

Find the expression for internal energy

To find the internal energy, we'll use the following equation: \\[ U = -\frac{\partial}{\partial \beta} \ln q_E \\] Where \(\beta = \frac{1}{kT}\) First, let's take the natural logarithm of the partition function and differentiate it with respect to \(\beta\): \\[ \ln q_E = \ln \left[m_0 \left(1 + \frac{m_1}{m_0} e^{-\varepsilon_1 \beta}\right)\right] \\] Let \(x = \frac{m_1}{m_0} e^{-\varepsilon_1 \beta}\), and differentiate regarding \(\beta\): \\[ \frac{\partial \ln q_E}{\partial \beta} = -\frac{m_1 \varepsilon_1 e^{-\varepsilon_1 \beta}}{m_0 + m_1 e^{-\varepsilon_1 \beta}} \\] Substitute this result back into the internal energy equation: \\[ U = -\left(-\frac{m_1 \varepsilon_1 e^{-\varepsilon_1 \beta}}{m_0 + m_1 e^{-\varepsilon_1 \beta}}\right) \\] For the limiting values of internal energy, we need to analyze as the temperature approaches zero and infinity: 1. As \(T \to 0\) or \(\beta \to \infty\), the exponential terms go to zero, and the internal energy becomes: \\[ U \to \frac{m_1 \varepsilon_1}{m_0} 0 = 0 \\] 2. As \(T \to \infty\) or \(\beta \to 0\), since both levels are populated, the internal energy will be: \\[ U = m_1 \varepsilon_1 \frac{1}{1 + \frac{m_0}{m_1}} = \frac{m_1}{m_0 + m_1} \varepsilon_1 \\] Thus, we have found the electronic partition function, the internal energy, and its limiting values as the temperature approaches zero and infinity.

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Key Concepts

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

Degeneracy in Quantum States
In quantum mechanics, degeneracy refers to the phenomenon where two or more different quantum states share the same energy level. This is a common occurrence in systems like atoms, molecules, and solids, where symmetries or identical particles lead to multiple states having equal energy.

For instance, consider an electron in an atom that can occupy different orbitals. When orbitals have the same energy due to the atom's symmetry, they are labeled as degenerate. The significance of degeneracy becomes evident when studying statistical thermodynamics, particularly when determining the electronic partition function of a system.

In the given exercise, the ground state with energy \(\varepsilon_{0}\) is \(m_{o}\)-fold degenerate, meaning there are \(m_{o}\) ways the ground state can be occupied without altering the energy of the system. Similarly, the first excited state at \(\varepsilon_{1}\) is \(m_{1}\)-fold degenerate. Degeneracy plays a key role in determining the properties of a system, such as its entropy and internal energy, as it influences the probable distributions of particles across energy levels.
Internal Energy of a System
The internal energy of a system in thermodynamics is the total energy contained within the system, attributed to the kinetic and potential energy of the particles. In quantum systems, this includes the energy levels that particles like electrons can occupy. The energy of these systems is quantized, meaning it can only exist in specific discrete values, often defined by quantum numbers.

Internal energy is affected by temperature changes and can be calculated from the partition function in statistical mechanics. As demonstrated in the exercise, the internal energy (\(U\)) can be expressed mathematically by taking the negative partial derivative of the natural logarithm of the electronic partition function (\(q_E\)) with respect to Beta (\(\beta=1/(kT)\)), where \(k\) is the Boltzmann constant and \(T\) is the temperature.

As temperature approaches zero, the internal energy will approach the energy of the ground state since higher energy states will be less populated. Conversely, at very high temperatures, the internal energy reflects the distribution of particles across all accessible states weighted by their energy levels.
Boltzmann Constant
The Boltzmann constant (\(k\)), named after the Austrian physicist Ludwig Boltzmann, is a fundamental physical constant that connects macroscopic and microscopic descriptions of nature. It plays a pivotal role in the field of statistical thermodynamics, where it relates the average kinetic energy of particles in a gas with the temperature of the gas.

The Boltzmann constant has units of energy per temperature (Joules per Kelvin) and is a key factor in the equation for the electronic partition function, where it appears in the exponent as the energy, epsilon (\(\varepsilon\)), divided by the product of \(k\) and the temperature (\(T\)). This exponential factor determines the probability of a system occupying a state with a certain energy level at a given temperature.

Understanding the role of the Boltzmann constant is essential when computing probabilities and energies in statistical mechanics. It allows the transition from individual particle behavior to the statistical average behavior of a system, providing a bridge between quantum mechanics and thermodynamics.
Statistical Thermodynamics
Statistical thermodynamics, or statistical mechanics, is the study of the relationship between the macroscopic properties of materials (such as temperature, volume, and pressure) and the microscopic behaviors of their constituent particles (like atoms, molecules, and electrons).

Central to this field is the concept of the partition function, which encapsulates all possible microstates of a system and their respective probabilities. It is the fundamental link from which thermodynamic properties can be derived. In the context of our exercise, the electronic partition function combines the principles of energy quantization, degeneracy, and distribution of particles across energy states according to the Boltzmann distribution.

By analyzing how a system's partition function behaves with temperature, as shown in the provided solution, one can determine important properties such as the internal energy. Statistical thermodynamics thus provides a powerful framework for predicting how a material's internal energy changes with temperature, offering deep insight into a wide range of physical phenomena.

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Most popular questions from this chapter

Calculate the internal energy of He, Ne, and Ar under standard thermodynamic conditions. Do you need to redo the entire calculation for each species?

Consider rotation about the C-C bond in ethane. A crude model for torsion about this bond is the "free rotor" model where rotation is considered unhindered. In this model the energy levels along the torsional degree of freedom are given by \\[ E_{j}=\frac{\hbar^{2} j^{2}}{2 I} \text { for } j=0,\pm 1,\pm 2, \ldots \\] In this expression \(I\) is the moment of inertia. Using these energies, the summation expression for the corresponding partition function is \\[ Q=\frac{1}{\sigma} \sum_{j=-\infty}^{\infty} e^{-E_{l} / k T} \\] where \(\sigma\) is the symmetry number. a. Assuming that the torsional degree of freedom is in the hightemperature limit, evaluate the previous expression for \(Q\) b. Determine the contribution of the torsional degree of freedom to the molar constant-volume heat capacity. c. The experimentally determined \(C_{v}\) for the torsional degree of freedom is approximately equal to \(\mathrm{R}\) at \(340 .\) K. Can you rationalize why the experimental value is greater that than predicted using the free rotor model?

The isotope exchange reaction for \(\mathrm{Cl}_{2}\) is as follows: \(^{35} \mathrm{Cl}^{35} \mathrm{Cl}+^{37} \mathrm{Cl}^{37} \mathrm{Cl} \rightleftharpoons 2^{37} \mathrm{Cl}^{35} \mathrm{Cl} .\) The equilibrium constant for this reaction is \(\sim 4 .\) Furthermore, the equilibrium constant for similar isotope-exchange reactions is also close to this value. Demonstrate why this would be so.

The four energy levels for atomic vanadium (V) have the following energies and degeneracies: $$\begin{array}{ccc} \text { Level }(\boldsymbol{n}) & \text { Energy }\left(\mathrm{cm}^{-1}\right) & \text {Degeneracy } \\ \hline 0 & 0 & 4 \\ 1 & 137.38 & 6 \\ 2 & 323.46 & 8 \\ 3 & 552.96 & 10 \end{array}$$ What is the contribution to the average molar energy from electronic degrees of freedom for \(\mathrm{V}\) when \(T=298 \mathrm{K} ?\)

In "Direct Measurement of the Size of the Helium Dimer" by F. Luo, C. F. Geise, and W. R. Gentry (J. Chemical Physics \(104[1996]: 1151\) ), evidence for the helium dimer is presented. As one can imagine, the chemical bond in the dimer is extremely weak, with an estimated value of only \(8.3 \mathrm{mJ} \mathrm{mol}^{-1}\) a. An estimate for the bond length of \(65 \AA\) is presented in the paper. Using this information, determine the rotational constant for \(\mathrm{He}_{2}\). Using this value for the rotational constant, determine the location of the first rotational state. If correct, you will determine that the first excited rotational level is well beyond the dissociation energy of He \(_{2}\) b. Consider the following equilibrium between He \(_{2}\) and its atomic constituents: \(\mathrm{He}_{2}(g) \rightleftharpoons 2 \mathrm{He}(g) .\) If there are no rotational or vibrational states to consider, the equilibrium is determined exclusively by the translational degrees of freedom and the dissociation energy of He \(_{2}\). Using the dissociation energy provided earlier and \(V=1000 . \mathrm{cm}^{3}\) determine \(K_{P}\) assuming that \(T=10 .\) K. The experiments were actually performed at \(1 \mathrm{mK} ;\) why was such a low temperature employed?

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