Question

A two-level system is characterized by an energy separation of \(1.30 \times 10^{-18} \mathrm{J}\). At what temperature will the population of the ground state be 5 times greater than that of the excited state?

Short answer

The population of the ground state will be 5 times greater than that of the excited state at a temperature of \(4.52 \times 10^4 K\).

Step-by-step solution

Write the Boltzmann distribution formula for the given system

The Boltzmann distribution formula for a two-level system is given by \[\frac{N_2}{N_1} = \frac{g_2}{g_1} \exp \left(-\frac{\Delta E}{k_B T}\right)\] where \(N_1\) and \(N_2\) are the populations of the ground and excited states, respectively, \(g_1\) and \(g_2\) are the degeneracies of the ground and excited states, \(\Delta E\) is the energy difference between the two states, \(k_B\) is the Boltzmann constant with a value of \(1.38 \times 10^{-23} JK^{-1}\), and \(T\) is the temperature in Kelvin.

Determine the condition based on the given information

We are given that the population of the ground state, \(N_1\), is 5 times greater than that of the excited state, \(N_2\). So, \[N_1 = 5 N_2\] Also, we can assume that the degeneracies of the ground and excited states are equal, i.e., \(g_1 = g_2\).

Substitute the given values into the Boltzmann distribution formula and solve for T

Since the degeneracies are equal, we can write the Boltzmann distribution formula as \[\frac{N_2}{N_1} = \exp\left(-\frac{\Delta E}{k_B T}\right)\] We have \(N_1 = 5 N_2\), so we can substitute this in the formula as \[\frac{1}{5} = \exp\left(-\frac{1.30 \times 10^{-18} J}{(1.38 \times 10^{-23} JK^{-1}) T}\right)\] Now, we need to solve for the temperature, T. To do this, we will take the natural logarithm of both sides of the equation. \[\ln\left(\frac{1}{5}\right) = \ln\left(\exp\left(-\frac{1.30 \times 10^{-18} J}{(1.38 \times 10^{-23} JK^{-1}) T}\right)\right) = -\frac{1.30 \times 10^{-18} J}{(1.38 \times 10^{-23} JK^{-1}) T}\] Next, we will isolate the temperature, T, by solving for it: \[T = -\frac{1.30 \times 10^{-18} J}{(1.38 \times 10^{-23} JK^{-1}) \ln\left(\frac{1}{5}\right)}\]

Calculate the temperature

Now, we can plug in the values and calculate the temperature: \[T = -\frac{1.30 \times 10^{-18} J}{(1.38 \times 10^{-23} JK^{-1}) \ln\left(\frac{1}{5}\right)} = 4.52 \times 10^4 K\] So, the population of the ground state will be 5 times greater than that of the excited state at a temperature of \(4.52 \times 10^4 K\).

Key concepts

Two-Level System

Imagine a two-level system as a simple model where we focus on atoms or molecules that can occupy two energy states: a lower energy state (ground state) and a higher energy state (excited state). This concept is fundamental in statistical mechanics and is used to explain various phenomena in physics and chemistry.
In a two-level system, the particle can only exist in one of these two states. The energy difference between the states is known as the energy separation. This difference determines how likely particles are to be found in the excited state as the temperature changes.
Understanding two-level systems helps us learn about how particles behave when they can only occupy a few energy states, which simplifies the study of more complex systems.

Energy Separation

Energy separation is the energy difference (\[\Delta E\]) between the ground state and the excited state of a system. In our exercise, this value is given as \(1.30 imes 10^{-18} \mathrm{J}\). This separation is crucial because it affects how particles distribute among the available states.
The concept is particularly important when examining temperature effects on particle populations. As the temperature rises, the likelihood of particles moving to an excited state may increase, depending on the energy separation. A smaller energy separation means it's easier for particles to transition to the higher energy state, while a larger energy separation requires more thermal energy for such transitions.
Overall, by understanding energy separation, we gain insight into how temperature can influence the behavior of particles in a two-level system.

Degeneracy

Degeneracy refers to the number of ways a particular energy state can be achieved. In simple terms, it counts the possible configurations that share the same energy level. For a two-level system, ground and excited states can have different degeneracies, denoted as \(g_1\) for the ground state and \(g_2\) for the excited state.
In the step by step solution given, it's assumed that the degeneracies are equal (\(g_1 = g_2\)). This assumption simplifies the usage of the Boltzmann distribution because the ratio of degeneracies becomes one, and we can focus on the exponential term in the formula. However, in real-world systems, degeneracies might differ, significantly affecting the population ratio between different energy states.
Understanding degeneracy helps explain why certain states are more or less populated and clarifies the impact of symmetry and quantum mechanics on a system's properties.

Population Ratio

The population ratio in a two-level system indicates how many particles are in each energy state. It's essentially the ratio \(\frac{N_2}{N_1}\), where \(N_1\) is the population of the ground state and \(N_2\) is the population of the excited state. The Boltzmann distribution describes this ratio mathematically.
In the context of our exercise, the population ratio is given by the condition that the population of the ground state (\(N_1\)) is five times that of the excited state (\(N_2\)). Thus, \(N_1 = 5N_2\), leading to a ratio of \(\frac{1}{5}\). The solution uses this information with the Boltzmann formula to calculate the temperature at which this population ratio holds.
Understanding population ratios allows us to determine how particles are distributed across energy levels and provides insights into the thermal behavior of various materials and systems.