Problem 130 - Chemistry An Atoms First Approach

Question

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and \(589.6 \mathrm{nm}\). What are the frequency and the energy of a photon of light at each of these wavelengths? What are the energies in \(\mathrm{kJ} / \mathrm{mol} ?\)

Step-by-Step Solution

Answer

The frequencies and energies of the two emission lines are: 1. Wavelength \(589.0\,\text{nm}\): frequency \(v_1\) = \(5.090 \times 10^{14}\,\mathrm{s}^{-1}\), energy \(E_1\) = \(3.3726 \times 10^{-19}\,\mathrm{J}\) or \(203.0\,\frac{\text{kJ}}{\text{mol}}\) 2. Wavelength \(589.6\,\text{nm}\): frequency \(v_2\) = \(5.083 \times 10^{14}\,\mathrm{s}^{-1}\), energy \(E_2\) = \(3.3695 \times 10^{-19}\,\mathrm{J}\) or \(202.7\,\frac{\text{kJ}}{\text{mol}}\)

Step 1: Convert wavelengths to meters

First, let's convert the given wavelengths to meters by multiplying with the appropriate conversion factor (1 nm = \(1 \times 10^{-9}\) m): Wavelength 1: \(589.0\,\text{nm}\times\frac{1 \times 10^{-9}m}{1 \text{nm}}= 5.89 \times 10^{-7} \mathrm{m}\) Wavelength 2: \(589.6\,\text{nm}\times\frac{1 \times 10^{-9}m}{1\text{nm}}= 5.896 \times 10^{-7} \mathrm{m}\)

Step 2: Calculate the frequency for each wavelength

Using the equation \(v = \frac{c}{\lambda}\), we find the frequency for each wavelength: Frequency 1: \(v_1 = \frac{2.998 \times10^{8}\,\text{m/s}}{5.89 \times 10^{-7}\,\text{m}} = 5.090 \times 10^{14}\; \mathrm{s}^{-1}\) Frequency 2: \(v_2 = \frac{2.998 \times10^{8}\,\text{m/s}}{5.896 \times 10^{-7}\,\text{m}} = 5.083 \times 10^{14}\; \mathrm{s}^{-1}\)

Step 3: Find the energy of a photon of light at each frequency

Using the Planck's equation (\(E = h v\)), we find the energy for each frequency: Energy 1: \(E_1 = (6.63 \times 10^{-34}\,\text{J·s}) \times (5.090 \times 10^{14}\,\text{s}^{-1}) = 3.3726 \times 10^{-19}\; \mathrm{J}\) Energy 2: \(E_2 = (6.63 \times 10^{-34}\,\text{J·s}) \times (5.083 \times 10^{14}\,\text{s}^{-1}) = 3.3695 \times 10^{-19}\; \mathrm{J}\)

Step 4: Convert energies to kJ/mol

We convert energies to kJ/mol using the Avogadro's number (\(6.022 \times 10^{23}\, \mathrm{mol}^{-1}\)) : Energy 1 in kJ/mol: \(\frac{3.3726 \times 10^{-19}\,\text{J}}{1 \,\text{photon}} \times \frac{1\,\text{kJ}}{10^3 \,\text{J}} \times\frac{6.022 \times 10^{23}\,\text{photons}}{1 \,\text{mol}} = 203.0\,\mathrm{kJ/mol}\) Energy 2 in kJ/mol: \(\frac{3.3695 \times 10^{-19}\,\text{J}}{1 \,\text{photon}} \times \frac{1\,\text{kJ}}{10^3 \,\text{J}} \times\frac{6.022 \times 10^{23}\,\text{photons}}{1 \,\text{mol}} = 202.7\,\mathrm{kJ/mol}\) Now we have the frequency and energy of a photon of light at each wavelength and their corresponding energies in kJ/mol: 1. Wavelength \(589.0\,\text{nm}\): frequency \(v_1\) = \(5.090 \times 10^{14}\,\mathrm{s}^{-1}\), energy \(E_1\) = \(3.3726 \times 10^{-19}\,\mathrm{J}\) or \(203.0\,\frac{\text{kJ}}{\text{mol}}\) 2. Wavelength \(589.6\,\text{nm}\): frequency \(v_2\) = \(5.083 \times 10^{14}\,\mathrm{s}^{-1}\), energy \(E_2\) = \(3.3695 \times 10^{-19}\,\mathrm{J}\) or \(202.7\,\frac{\text{kJ}}{\text{mol}}\)

Key Concepts

Sodium Vapor Lamp Emission

When we discuss sodium vapor lamp emission, we're delving into a common type of lighting that's especially noted for its high efficiency and distinctive yellow glow. These lamps operate by passing an electric current through vaporized sodium metal, which—in a process of atomic excitement and relaxation—emits light at particular wavelengths, most notably at around 589.0 nm and 589.6 nm.

These specific wavelengths result in the lamp's characteristic bright yellow light, which can be quite piercing due to its narrow spectral bandwidth. These emissions are a result of sodium electron transitions, and understanding the photon energies associated with them brings us to Planck's equation.

Planck's Equation

Moving on to Planck's equation, it's fundamental to quantum mechanics and is pivotal to understanding how energy levels of photons relate to their frequency. Planck's equation is succinctly stated as \( E = h u \), where \( E \) represents the energy of a photon, \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) J·s), and \( u \) is the frequency of the radiation.

With this equation, we can calculate the energy of the photons emitted by a sodium vapor lamp. It's a bridge between the classical wave description of light and the concept of light as particles or 'quanta'. In the case of the sodium vapor lamp, knowing the frequencies of the light emissions allows us to use Planck's equation to find the exact energy of each photon.

Frequency and Wavelength Relationship

The frequency of light and its wavelength are inversely proportional to each other, as guided by the equation \( c = \lambda u \), where \( c \) is the speed of light in a vacuum (approximately \( 3.00 \times 10^8 \) m/s), \( \lambda \) is the wavelength, and \( u \) is the frequency. This relationship, also called the wave equation, helps us translate the wavelength of light emitted from a sodium vapor lamp into its frequency.

It's crucial to know this because Planck's equation uses frequency to determine energy. By converting the given wavelengths of 589.0 nm and 589.6 nm to frequencies, we can then employ Planck's equation to find the energies of the photons associated with these wavelengths. This cross-disciplinary knowledge highlights the interconnection between different properties of light.

Avogadro's Number

Lastly, Avogadro's number plays a significant role in chemistry and physics by providing a link between the microcosm of atoms and molecules and the macro world we live in. It is defined as the number of constituent particles (usually atoms or molecules) that are contained in one mole of a substance. Avogadro's number is a huge value, approximately \( 6.022 \times 10^{23} \) entities per mole.

This number allows us to scale up individual photon energies to the molar level, thus converting from energy per photon to energy per mole, a conversion that's much more pertinent to chemical reactions and processes. In the context of our sodium vapor lamp emission problem, after finding the energy of a single photon, we used Avogadro's number to calculate the energy per mole in kJ/mol, which is more relevant for chemists and can be directly related to macroscopic quantities.