Question

Assume that a system has a very large number of energy levels given by the formula \(\varepsilon=\varepsilon_{0} l^{2}\) with \(\varepsilon_{0}=\) \(1.75 \times 10^{-22} \mathrm{J},\) where \(l\) takes on the integral values 1,2 \(3, \ldots .\) Assume further that the degeneracy of a level is given by \(g_{l}=2 l .\) Calculate the ratios \(n_{4} / n_{1}\) and \(n_{8} / n_{1}\) for \(T=125 \mathrm{K}\) and \(T=750 . \mathrm{K}\)

Short answer

For a system with energy levels given by \(\varepsilon = \varepsilon_0 l^2\) and degeneracy formula \(g_l = 2l\), the ratios of the number of particles in energy levels \(n_4/n_1\) and \(n_8/n_1\) are calculated using the Boltzmann distribution. At \(T = 125\mathrm{K}\), the ratios are approximately: \(\frac{n_4}{n_1} \approx 6.25 \times 10^{-2}\) and \(\frac{n_8}{n_1} \approx 5.03 \times 10^{-9}\). While at \(T = 750\mathrm{K}\), the ratios are approximately: \(\frac{n_4}{n_1} \approx 0.31\) and \(\frac{n_8}{n_1} \approx 1.13 \times 10^{-3}\).

Step-by-step solution

Calculate the energy levels for l=1, 4, 8

Using the given energy formula \(\varepsilon = \varepsilon_0 l^2\), find the energy levels for \(l = 1, 4, 8\) with \(\varepsilon_0 = 1.75 \times 10^{-22}\mathrm{J}\). \[ \varepsilon_1 = \varepsilon_0 (1)^2 = 1.75 \times 10^{-22}\mathrm{J}, \] \[ \varepsilon_4 = \varepsilon_0 (4)^2 = 2.8 \times 10^{-21}\mathrm{J}, \] \[ \varepsilon_8 = \varepsilon_0 (8)^2 = 1.12 \times 10^{-20}\mathrm{J}. \]

Calculate the ratios using Boltzmann distribution

Using Boltzmann distribution, the ratio \(n_4/n_1\) and \(n_8/n_1\) will be: \[ \frac{n_4}{n_1} = \frac{g_4e^{-\beta \varepsilon_4}}{g_1e^{-\beta \varepsilon_1}} \] \[ \frac{n_8}{n_1} = \frac{g_8e^{-\beta \varepsilon_8}}{g_1e^{-\beta \varepsilon_1}} \]

Perform the calculations for both temperatures

For \(T = 125\mathrm{K}\) and \(T = 750\mathrm{K}\), we can use the given degeneracy formula \(g_l = 2l\) and above energy levels (\(\varepsilon_l\)) to calculate the ratios. For \(T = 125\mathrm{K}\): \(\beta\) = \(\frac{1}{k_BT}\) = \(\frac{1}{1.38\times10^{-23}\cdot1.25\times10^2}\) Using \(\beta\), we get: \[ \frac{n_4}{n_1} = \frac{2\cdot4e^{-\beta (2.8 \times 10^{-21})}}{2\cdot1e^{-\beta (1.75 \times 10^{-22})}} \approx 6.25 \times 10^{-2} \] \[ \frac{n_8}{n_1} = \frac{2\cdot8e^{-\beta (1.12 \times 10^{-20})}}{2\cdot1e^{-\beta (1.75 \times 10^{-22})}} \approx 5.03 \times 10^{-9} \] For \(T = 750\mathrm{K}\): \(\beta\) = \(\frac{1}{k_BT}\) = \(\frac{1}{1.38\times10^{-23}\cdot7.5\times10^2}\) Using \(\beta\), we get: \[ \frac{n_4}{n_1} = \frac{2\cdot4e^{-\beta (2.8 \times 10^{-21})}}{2\cdot1e^{-\beta (1.75 \times 10^{-22})}} \approx 0.31 \] \[ \frac{n_8}{n_1} = \frac{2\cdot8e^{-\beta (1.12 \times 10^{-20})}}{2\cdot1e^{-\beta (1.75 \times 10^{-22})}} \approx 1.13 \times 10^{-3} \] So, at \(T = 125\mathrm{K}\), the ratios are approximately: \[ \frac{n_4}{n_1} \approx 6.25 \times 10^{-2}, \;\;\;\; \frac{n_8}{n_1} \approx 5.03 \times 10^{-9} \] and at \(T = 750\mathrm{K}\), the ratios are approximately: \[ \frac{n_4}{n_1} \approx 0.31, \;\;\;\; \frac{n_8}{n_1} \approx 1.13 \times 10^{-3} \]

Key concepts

Energy Levels

In physics and chemistry, energy levels refer to the specific energies that electrons can have within an atom or a molecule. In the context of the problem we're discussing, energy levels are defined for a system using a mathematical formula. Here, the energy of a level is given by \( \varepsilon = \varepsilon_{0} l^2 \), where \( l \) represents an integer sequence such as 1, 2, 3, and so on. This sequence indicates different quantized states or levels that a system's electrons or particles can occupy.

- **Energy levels are quantized**, meaning they have discrete rather than continuous values.- **In this model**, the energy increases with the square of the quantum level, \( l \). Thus, when \( l \) increases, \( \varepsilon \) increases significantly.- **The base energy level** is given as \( \varepsilon_0 = 1.75 \, \times \, 10^{-22} \, \text{J} \), which serves as a scaling factor for calculating other energy levels.

Understanding how energy levels are derived allows us to calculate properties of the system, such as its capacity to store and release energy, and the manner in which particles transition between different states under various conditions.

Degeneracy

Degeneracy in the context of energy levels refers to the number of distinct states or configurations that share the same energy. More plainly, it's how many ways the same energy level can occur due to the system's spatial or symmetrical properties. In our case, degeneracy \( g_l \) for each level \( l \) is defined as \( 2l \).

- **Degeneracy provides a measure** of how 'occupied' an energy level can be with particles or electrons.- **The formula \( g_l = 2l \)** highlights that the degeneracy of level \( l \) is directly proportional to \( l \). When \( l \) is larger, the number of possible states at that energy level also increases.- **Understanding degeneracy** is vital for predicting how a system's particles will distribute among available energy levels.

This concept of degeneracy affects calculations in statistical mechanics, where it's used to determine, among many other things, the probability of a system at a certain energy state during thermal distribution calculations.

Thermal Population

Thermal population concerns how particles occupy the available energy levels in a system at thermal equilibrium. According to the Boltzmann distribution, the distribution of particles across energy levels is based on the temperature of the system and the energy differences between levels.

- **Boltzmann Distribution Formula:** This is mathematically described by the expression \( n_l \propto g_l \cdot e^{-\beta \varepsilon_l} \), where \( n_l \) denotes the number of particles at level \( l \), \( g_l \) is the degeneracy, and \( \beta = 1/(k_B T) \) with \( k_B \) being the Boltzmann constant and \( T \) the temperature.- **At higher temperatures**, particles are more likely to be found in higher energy states because thermal energy overcomes the energy differences.- **At lower temperatures**, particles tend to settle in lower energy states due to less thermal agitation.- **Importance of this concept**: By understanding thermal population, we know how likely a particle is to be in a specific energy state, which is crucial for predicting and analyzing a system's macroscopic properties like pressure or heat capacity.

This topic forms a core part of statistical mechanics, explaining how microscopic laws affect macroscopic observations.