Calculate the relative Boltzmann population of the v = 1 and v = 0 vibrational energy levels of a diatomic molecule at 25°C if they are seperated by 1000 cm^-1

To calculate the relative Boltzmann population between the \( v = 1 \) and \( v = 0 \) vibrational energy levels of a diatomic molecule, we use the **Boltzmann distribution**:

\[
\frac{N_1}{N_0} = e^{-\frac{\Delta E}{k_B T}}
\]

Where:
- \( \frac{N_1}{N_0} \) is the relative population of the \( v = 1 \) and \( v = 0 \) energy levels.
- \( \Delta E \) is the energy difference between the levels.
- \( k_B \) is the Boltzmann constant (\( 1.3807 \times 10^{-23} \, \text{J/K} \)).
- \( T \) is the temperature in Kelvin (25°C = 298.15 K).

The energy difference between vibrational levels is often given in terms of wavenumbers \( \tilde{\nu} \) (in \( \text{cm}^{-1} \)). To convert wavenumbers to energy, use the relation:

\[
\Delta E = h c \tilde{\nu}
\]

Where:
- \( h \) is Planck's constant (\( 6.6261 \times 10^{-34} \, \text{J·s} \)).
- \( c \) is the speed of light (\( 3.00 \times 10^{10} \, \text{cm/s} \)).
- \( \tilde{\nu} \) is the wavenumber separation, which is \( 1000 \, \text{cm}^{-1} \) in this case.

Now, let's calculate \( \Delta E \), convert it to joules, and then calculate the Boltzmann ratio.

\[
\Delta E = (6.6261 \times 10^{-34} \, \text{J·s}) \times (3.00 \times 10^{10} \, \text{cm/s}) \times (1000 \, \text{cm}^{-1})
\]
\[
\Delta E = 1.9878 \times 10^{-20} \, \text{J}
\]

Now apply the Boltzmann formula:

\[
\frac{N_1}{N_0} = e^{-\frac{1.9878 \times 10^{-20} \, \text{J}}{(1.3807 \times 10^{-23} \, \text{J/K}) \times 298.15 \, \text{K}}}
\]
\[
\frac{N_1}{N_0} = e^{-\frac{1.9878 \times 10^{-20}}{4.117 \times 10^{-21}}}
\]
\[
\frac{N_1}{N_0} = e^{-4.83}
\]
\[
\frac{N_1}{N_0} \approx 0.0079
\]

Thus, the relative population of the \( v = 1 \) level compared to the \( v = 0 \) level is approximately 0.0079, meaning that for every 1000 molecules in the \( v = 0 \) level, about 8 are in the \( v = 1 \) level.

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Would you like further details on any part of the calculation? 

Here are 5 related questions:
1. How does increasing temperature affect the relative population of vibrational levels?
2. What is the significance of the wavenumber in vibrational spectroscopy?
3. How would the relative population change if the energy level separation was 1500 cm\(^{-1}\)?
4. What role does the Boltzmann constant play in thermal distributions?
5. How does the vibrational energy level distribution compare with rotational energy levels?

**Tip**: The Boltzmann distribution highlights that higher energy levels become more populated at higher temperatures.

Calculate the relative Boltzmann population of the v = 1 and v = 0 vibrational energy levels of a diatomic molecule at 25°C if they are separated by 1000 cm^-1

This detailed solution covers the calculation of the relative Boltzmann population between the v = 1 and v = 0 vibrational energy levels of a diatomic molecule at 25°C. Using the Boltzmann distribution formula and key physical constants, the energy separation is determined, and the population ratio is calculated as approximately 0.0079. Suitable for undergraduate students studying thermodynamics or statistical mechanics.

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