Question

The vibrational frequency of \(\mathrm{I}_{2}\) is \(208 \mathrm{cm}^{-1}\). At what temperature will the population in the first excited state be half that of the ground state?

Short answer

The temperature at which the population in the first excited state of I₂ is half that of the ground state is approximately 707 K.

Step-by-step solution

Convert the vibrational frequency to energy

First, we need to convert the vibrational frequency from wavenumbers to energy. We can do this using the following formula: \[E=\nu hc\] where \(E\) is the energy, \(\nu\) is the frequency given in cm⁻¹, \(h = 6.626\times 10^{-34} \ \mathrm{J\cdot s}\), and \(c = 2.998\times10^{10} \ \mathrm{cm\cdot s^{-1}}\).

Calculate the energy difference

Now, we can calculate the energy difference between the ground state and the first excited state: \[\Delta E = E_{1} - E_{0} = E\] The energy difference is equal to the energy we calculated in step 1.

Apply the Boltzmann distribution

The Boltzmann distribution gives the relative populations of different energy levels: \[\frac{N_{1}}{N_{0}} = e^{-\frac{\Delta E }{kT}}\] where \(\frac{N_{1}}{N_{0}}\) is the ratio of populations in the excited state to that in the ground state, \(\Delta E\) is the energy difference between the states, \(k = 1.38\times10^{-23} \ \mathrm{J\cdot K^{-1}}\) is the Boltzmann constant, and \(T\) is the temperature.

Set the population ratio

We are given that the population in the first excited state is half that of the ground state. So, the ratio of the populations is: \[\frac{N_{1}}{N_{0}} = \frac{1}{2}\]

Solve for temperature

Substitute the population ratio and the energy difference into the Boltzmann distribution formula and solve for the temperature, \(T\): \[\frac{1}{2} = e^{-\frac{\Delta E }{kT}}\] \[T = -\frac{\Delta E}{k \ln(\frac{1}{2})}\] Plug in the values for \(\Delta E\) and \(k\) we obtained in previous steps, and solve for \(T\): \[T \approx \frac{(208 \ \mathrm{cm}^{-1})(6.626\times 10^{-34}\ \mathrm{J\cdot s})(2.998\times10^{10} \ \mathrm{cm\cdot s^{-1}})}{1.38\times10^{-23} \ \mathrm{J\cdot K^{-1}} \ln(2)} \approx 706.5 \ \mathrm{K}\]

Round to the nearest whole number

The temperature where the population in the first excited state is half of that in the ground state is approximately 707 K when rounded to the nearest whole number.

Key concepts

Vibrational Frequency

Vibrational frequency is the rate at which atoms in a molecule move as they vibrate. It's a critical aspect of molecular physics because it influences the energy levels of molecules. The energy associated with the vibrational frequency can be converted from wavenumbers (cm⁻¹), a common unit in spectroscopy, to joules using the formula \(E=u hc\), where \(E\) represents the energy, \(u\) is the vibrational frequency, \(h\) is Planck's constant, and \(c\) is the speed of light.

In simplifying the understanding of this conversion, think of it as translating the rhythmic movement frequency into a quantifiable energy form. This is much like knowing how many energy calories are in food by converting the nutritional values into something we can measure and understand better.

Excited State Population

The excited state population in a molecule refers to the number of particles that have absorbed energy and moved from a lower energy level, or the ground state, to a higher energy 'excited' state. This concept is essential in understanding molecules' behavior under various energy influences. Using the Boltzmann distribution, we can determine this population's ratio in comparison to the ground state.

For instance, the exercise asks at what temperature the population of iodine in the first excited state will be half of that in the ground state. We rely on the Boltzmann distribution formula \(\frac{N_{1}}{N_{0}} = e^{-\frac{\Delta E }{kT}}\) to find this temperature, where \(\frac{N_{1}}{N_{0}}\) is the population ratio, \(\Delta E\) is the energy difference, \(k\) is the Boltzmann constant, and \(T\) is the temperature.

Energy Difference Calculation

The calculation of energy difference between molecular energy levels is pivotal in understanding transitions such as absorption and emission. For the given molecule of iodine (\(\mathrm{I}_{2}\)), the energy difference \(\Delta E\) between the ground state and the first excited state is equal to the energy calculated using the frequency of vibration and Planck's equation.

Through a simple formula, \(E=u hc\), we've established how to find the energy from a vibration frequency. By understanding the transition from the ground state to the first excited state as a single quantum energy jump, we can see that \(\Delta E\) would be the same as the energy calculated for a single vibrational quantum.

Temperature and Energy Level Population Relationship

Temperature plays a significant role in determining the distribution of particles among various energy levels in a system. A higher temperature generally increases the population of excited states due to the greater availability of thermal energy. In our iodine example, we see this relationship through the Boltzmann distribution, which indicates that the excited state population depends exponentially on the ratio of energy difference to the product of Boltzmann constant and temperature \(\frac{\Delta E}{kT}\).

To further elucidate this relationship, by rearranging the Boltzmann equation to solve for temperature when given a specific population ratio, we can calculate the temperature at which a certain proportion of the particles will be in an excited state. This vividly shows how intimately entwined temperature is to the behavior and distribution of particles within different energy levels.