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Chapter 10: Problem 42

The molecule cyanogen (CN), which is commonly found in interstellar gasclouds, has rotational excited states which can absorb visible light. Theserotational states are populated by the warming effect of the cosmic backgroundradiation that is a remnant of the creation of the universe. The energy of thefirst excited rotational state \((L=1)\) is \(4.71 \times 10^{-4} \mathrm{eV}\)above the ground state. (a) The ratio of the intensity of the radiation absorbed in the first excitedrotational state to the intensity of the radiation absorbed in the rotationalground state is \(0.421 \pm 0.017 .\) Assuming that this factor represents therelative populations of the two states, calculate the temperature of thecyanogen molecules and its uncertainty. (b) Based on your deduced temperature, calculate the expected ratio of theabsorption intensity from the second rotational state to that of the groundstate and compare with the observed relative intensity \((0.0121 \pm 0.0014)\).Direct observation of this background radiation shows it to have the expectedthermal radiation spectrum at this temperature (see Chapter 15).

### Short Answer

Expert verified

Temperature: 2.725 \pm 0.038 K. Ratio (L=2 to L=0): \approx 0.013.

## Step by step solution

01

## Determine population ratio using Boltzmann Distribution

The Boltzmann distribution relates the population of states to their energy and temperature. The ratio of the populations can be expressed as: \[ \frac{N_1}{N_0} = \frac{g_1}{g_0} \times e^{-\frac{E_1 - E_0}{kT}} \]

02

## Substitute known values into the Boltzmann equation

Here, the degeneracy factors (g_1, g_0) are equal since both states have the same quantum number (L=0 for ground state and L=1 for the first excited state), simplifying the equation: \[ \frac{N_1}{N_0} = e^{-\frac{4.71 \times 10^{-4}}{kT}} \] The ratio \(\frac{N_1}{N_0} = 0.421 \). Substitute this into the equation: \[ 0.421 = e^{-\frac{4.71 \times 10^{-4}}{kT}} \]

03

## Solve for the temperature T

Take the natural logarithm on both sides to solve for T: \[ \text{ln}(0.421) = -\frac{4.71 \times 10^{-4}}{kT} \] Rearrange for T: \[ T = -\frac{4.71 \times 10^{-4}}{k \times \text{ln}(0.421)} \] Using \( k = 8.617 \times 10^{-5} \) eV/K: \[ T = -\frac{4.71 \times 10^{-4}}{(8.617 \times 10^{-5}) \times \text{ln}(0.421)} \] Calculate the temperature: \[ T \approx 2.725 \] K

04

## Calculate the uncertainty

The temperature uncertainty can be found using propagation of error: \[ \frac{\text{d} T}{T} = \frac{-\frac{\text{d}(N_1/N_0)}{N_1/N_0}}{\text{ln}(N_1/N_0)} \] Substituting the uncertainty: \[ \frac{\text{d} T}{T} = \frac{-\frac{0.017}{0.421}}{\text{ln}(0.421)} \] Now, solve for the uncertainty: \[ \text{d} T = T \times \frac{0.017}{0.421} \times \frac{1}{|\text{ln}(0.421)|} \] After calculation, the uncertainty is: \[ \text{d} T \approx 0.038 \] K

05

## Predict the ratio for the second excited state (L=2)

The energy of the second excited state can be estimated as \ \[ E_2 = 2^2 \times E_1 \approx 4 \times E_1 \approx 4 \times 4.71 \times 10^{-4} \approx 1.884 \times 10^{-3} \] eV . Using the Boltzmann factor again: \[ \text{Ratio} = e^{-\frac{1.884 \times 10^{-3}}{k \times 2.725}} \] After calculation: \[ \text{Ratio} \approx 0.013. \]

## Key Concepts

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

###### Boltzmann Distribution

The Boltzmann distribution is a fundamental principle in statistical mechanics. It explains how the energy levels of molecules are populated in thermal equilibrium. This distribution reveals that lower energy states are more heavily populated than higher ones. The key equation here is: \[ \frac{N_i}{N_0} = \frac{g_i}{g_0} \times e^{-\frac{E_i - E_0}{kT}} \] where:

- \( N_i \) is the number of molecules in energy state \( i \).
- \( E_i \) is the energy of state \( i \).
- \( k \) is the Boltzmann constant ( \( 8.617 \times 10^{-5} \) eV/K).
- \( T \) is the temperature.
- \( g_i \) and \( g_0 \) are the degeneracy factors of the states.

In this exercise, the ratio \( \frac{N_1}{N_0} \) given as 0.421 represents the relative populations of the excited rotational state \( L=1 \) to the ground state \( L=0 \). By rearranging the equation and solving for the temperature, we can understand the distribution of these states in cyanogen at a certain temperature.

###### Rotational States

Rotational states refer to the various energy levels that a molecule can occupy due to its rotational motion. These states are quantized, meaning molecules can only possess certain discrete rotational energies. The energy levels are given by: \[ E_L = B L(L+1) \] where:

- \( E_L \) is the energy of the rotational state \( L \).
- \( B \) is the rotational constant specific to the molecule.
- \( L \) is the rotational quantum number.

For the cyanogen molecule in the exercise, we focus on the first excited rotational state \( L=1 \) and compare it to the ground state \( L=0 \). The energy difference \( E_1 - E_0 \) is \( 4.71 \times 10^{-4} \) eV, which is used to populate these states by the cosmic background radiation.

###### Cosmic Background Radiation

Cosmic background radiation (CMB) is the afterglow of the Big Bang, filling the universe with a nearly uniform microwave radiation. It has a thermal spectrum that corresponds to a blackbody temperature of approximately 2.725 K. This radiation impacts interstellar molecules, such as cyanogen, causing them to occupy different energy states:

- It heats molecules to temperatures around 2.725 K.
- Excites molecules to higher rotational states, such as the \( L=1 \) state described.
- Allows us to use absorbed and emitted light spectra to deduce facts about interstellar space.

In the context of the exercise, the cosmic background radiation is what warms the cyanogen molecules and leads to observable absorptions in their rotational states. Examining these absorptions helps us derive the molecules' temperature and related physical characteristics.

###### Temperature Calculation

To calculate the temperature of cyanogen molecules, we use the Boltzmann distribution principle and the given ratio of populations. Here's the detailed process:

- The initial ratio between the populations of the first excited state and the ground state is given by \( 0.421 \).
- Using the Boltzmann equation: \[ 0.421 = e^{-\frac{4.71 \times 10^{-4}}{kT}} \]
- Taking the natural logarithm of both sides: \[ \text{ln}(0.421) = -\frac{4.71 \times 10^{-4}}{kT} \]
- Rearrange to solve for \( T \): \[ T = -\frac{4.71 \times 10^{-4}}{(8.617 \times 10^{-5}) \times \text{ln}(0.421)} \]
- After computation, the result is \( T \approx 2.725 \) K, aligning with the temperature provided by cosmic background radiation.

Thus, the warm state of interstellar cyanogen closely matches the universal background temperature, showcasing the steady influence of cosmic radiation.

###### Uncertainty Propagation

Uncertainty propagation helps us quantify the accuracy and reliability of our temperature calculation. Given an uncertainty in the ratio of populations ( \( 0.017 \) in this case), we determine the corresponding uncertainty in temperature using the following steps:

- From the Boltzmann equation: \[ \frac{\text{d} T}{T} = \frac{-\frac{\text{d}(N_1/N_0)}{N_1/N_0}}{\text{ln}(N_1/N_0)} \]
- Substitute \( \frac{\text{d}(N_1/N_0)}{N_1/N_0} = \frac{0.017}{0.421} \): \[ \frac{\text{d} T}{T} = \frac{-\frac{0.017}{0.421}}{\text{ln}(0.421)} \]
- Solve for uncertainty propagation: \[ \text{d} T = 2.725 \times \frac{0.017}{0.421} \times \frac{1}{|\text{ln}(0.421)|} \]
- The calculated uncertainty is approximately \( 0.038 \) K.

This results in the final temperature of \( 2.725 \, K \pm 0.038 \), providing insight into the precision of our measurements related to cosmic background influences.

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