Problem 26 - Analytical Chemistry

Question

The transition for the cadmium 228.8 -nm line is a \(1 S_{0}-1 S_{1}\) transition. Calculate the ratio of \(N_{e} / N_{0}\) in an air-acetylene flame. What percent of the atoms is in the excited state? The velocity of light is \(3.00 \times 10^{10} \mathrm{~cm} / \mathrm{s}\), Planck's constant is \(6.62 \times 10^{-27}\) erg-s, and the Boltzmann constant is \(1.380 \times 10^{-16}\) erg \(\mathrm{K}^{-1}\)

Step-by-Step Solution

Answer

Approximately 8.90 x 10^-9% of the atoms are in the excited state.

Step 1: Determine the Energy

First, calculate the energy of the transition using the formula \[ E = \frac{hc}{\lambda} \]where \(h\) is Planck's constant \(6.62 \times 10^{-27} \text{ erg-s}\), \(c\) is the speed of light \(3.00 \times 10^{10} \text{ cm/s}\), and \(\lambda\) is the wavelength \(228.8 \text{ nm} = 228.8 \times 10^{-7} \text{ cm}\). Substituting these values, we get:\[ E = \frac{(6.62 \times 10^{-27} \text{ erg-s})(3.00 \times 10^{10} \text{ cm/s})}{228.8 \times 10^{-7} \text{ cm}} \approx 8.68 \times 10^{-12} \text{ erg} \]

Step 2: Apply Boltzmann Distribution

Use the Boltzmann distribution formula to find the ratio \(\frac{N_e}{N_0}\):\[\frac{N_e}{N_0} = e^{-E/(kT)}\]where \(E\) is the energy found in Step 1 \(8.68 \times 10^{-12} \text{ erg}\), \(k\) is the Boltzmann constant \(1.380 \times 10^{-16} \text{ erg/K}\), and \(T\) is the temperature of the air-acetylene flame. We assume a typical temperature \(T\) of \(2700 \text{ K}\):\[\frac{N_e}{N_0} = e^{-\frac{8.68 \times 10^{-12}}{1.380 \times 10^{-16} \times 2700}} \approx e^{-23.10} \approx 8.90 \times 10^{-11}\]

Step 3: Calculate Percentage of Excited Atoms

To find the percentage of atoms in the excited state, calculate:\[\text{Percentage} = \left(\frac{N_e}{N_0 + N_e}\right) \times 100\%\]Since \(N_e\) is much smaller than \(N_0\), \(N_0 + N_e \approx N_0\), thus:\[\text{Percentage} \approx \left(\frac{N_e}{N_0}\right) \times 100\% \approx 8.90 \times 10^{-11} \times 100\% = 8.90 \times 10^{-9}\%\]

Key Concepts

Energy Calculation

In order to understand the concept of energy calculation in atomic transitions, we start by calculating the energy emitted or absorbed during the transition. This can be computed using a known formula derived from quantum mechanics, which relates energy (E) to the frequency or wavelength of the light associated with the transition.
To begin, we use the equation \( E = \frac{hc}{\lambda} \) where **E** is the energy of the transition, **h** is Planck's constant \(6.62 \times 10^{-27} \text{ erg-s}\), **c** is the speed of light \(3.00 \times 10^{10} \text{ cm/s}\), and \(\lambda\) is the wavelength of the transition.
In this exercise, for the cadmium transition with a wavelength of 228.8 nm, we first convert the wavelength from nanometers to centimeters because the constant values are expressed in cgs units. This gives us 228.8 nm = \( 228.8 \times 10^{-7} \text{ cm} \). Using these values in the equation:

• Energy formula: \( E = \frac{(6.62 \times 10^{-27} \text{ erg-s})(3.00 \times 10^{10} \text{ cm/s})}{228.8 \times 10^{-7} \text{ cm}} \)
• This results in \( E \approx 8.68 \times 10^{-12} \text{ erg} \)

This energy value represents the change in energy associated with the transition, important for understanding how atomic states shift during the excitation process.

Atomic Transition

Atomic transition refers to the movement of an electron from one energy level to another within an atom. In the context of the cadmium exercise, we consider the transition labeled as a \(1 S_{0}-1 S_{1}\) transition. Here, the electron moves from an initial lower energy state to a higher energy excited state.
These transitions are quantized, meaning that they occur at specific energy levels, determined by the discrete energy states available to electrons within an atom. The energy calculated earlier is the difference between these states. The emitted or absorbed energy corresponds to specific electromagnetic radiation, often visible as light of a certain wavelength.
This process is modeled by the Boltzmann distribution, which gives us the ratio of the number of atoms in the excited state \(N_e\) to those in the ground state \(N_0\). By applying the Boltzmann formula:

• \(\frac{N_e}{N_0} = e^{-E/(kT)}\)

let us analyze how temperature affects this ratio. Given a typical flame temperature of \(2700 \text{ K}\) and using the calculated energy and constants, the calculated ratio becomes extremely small, \( \approx 8.90 \times 10^{-11} \). This reflects that under such conditions, only a negligible number of atoms are excited, as expected in thermal environments.

Excited State Percentage

The calculation of the percentage of atoms in the excited state is often necessary to understand the impact of excitation on the atom population. It effectively tells us how many atoms have absorbed enough energy to move up from the ground state to the excited state.
To find the percentage, the equation used is:

• \( \text{Percentage} = \left(\frac{N_e}{N_0 + N_e}\right) \times 100\% \)

Because \(N_e\), or the number of excited atoms, is much smaller than \(N_0\), the simplification \(N_0 + N_e \approx N_0\) is justified. This simplification leads to the approximate percentage formula:

• \( \text{Percentage} \approx \left(\frac{N_e}{N_0}\right) \times 100\% \)

Plugging in the values obtained from our Boltzmann ratio calculations, we find that fewer than one in a billion atoms, approximately \( 8.90 \times 10^{-9}\% \), are in the excited state at the given temperature condition. This minute percentage underscores the energy required to excite atoms in such environments, and reinforces the importance of temperature within atomic transitions.