In 1814 Fraunhofer observed a series of dark lines in the sun's spectrum, which he labeled A through H. About 50 years later, Gustav Kirchhoff noticed that the wavelength of light given off when sodium salts are added to a flame is the same as the wavelength of the D line in Fraunhofer's spectrum. He concluded that certain substances give off light when heated that has the same wavelength as the light absorbed under other conditions. The wavelength of the characteristic yellow-orange light emitted by sodium ions in a burner flame is 589.5923 nm. What is the energy of this light, in units of kJ/mol?

Question

Quizplus · Accepted Answer

To find the energy of the light in kJ/mol, we can use the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s), $c$ is the speed of light ($3.00 \times 10^8$ m/s), and $\lambda$ is the wavelength in meters. First, convert the wavelength from nm to m: $589.5923 \, \text{nm} = 589.5923 \times 10^{-9} \, \text{m}$. Then, calculate the energy per photon: $E = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{589.5923 \times 10^{-9} \, \text{m}}$. This gives $E = 3.37 \times 10^{-19} \, \text{J/photon}$. To convert to kJ/mol, multiply by Avogadro's number ($6.022 \times 10^{23} \, \text{mol}^{-1}$) and convert J to kJ: $E = 3.37 \times 10^{-19} \, \text{J/photon} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \times 10^{-3} \, \text{kJ/J} = 202.9 \, \text{kJ/mol}$, which falls in the range of 100 to 1000 kJ/mol.

In 1814 Fraunhofer Observed a Series of Dark Lines in the Sun's