To determine the energy of a photon emitted by a sodium vapor lamp, we use the fundamental relationship between a photon's energy, its wavelength, Planck's constant, and the speed of light. This relationship is a cornerstone of quantum physics and explains how light carries energy.

Photon Energy Formula

The energy \(E\) of a photon can be calculated using the formula:

\[E = \frac{hc}{\lambda}\]

Where:

• \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \text{ J}\cdot\text{s}\).
• \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^8 \text{ m/s}\).
• \(\lambda\) is the wavelength of the photon.

Wavelength Conversion

The given wavelength of the photon from the sodium vapor lamp is \(5890 \text{ Å}\) (Ångströms). To use it in our formula with standard units, we must convert Ångströms to meters, as \(1 \text{ Å} = 10^{-10} \text{ m}\).

So, the wavelength \(\lambda\) in meters is:

\[\lambda = 5890 \text{ Å} \times \frac{10^{-10} \text{ m}}{1 \text{ Å}} = 5890 \times 10^{-10} \text{ m} = 5.89 \times 10^{-7} \text{ m}\]

Energy Calculation in Joules

Now, we can substitute the values of \(h\), \(c\), and \(\lambda\) into the photon energy formula to find the energy in Joules (J):

\[E = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (3 \times 10^8 \text{ m/s})}{5.89 \times 10^{-7} \text{ m}}\]

\[E = \frac{19.878 \times 10^{-26} \text{ J}\cdot\text{m}}{5.89 \times 10^{-7} \text{ m}}\]

\[E \approx 3.3748 \times 10^{-19} \text{ J}\]

Energy Conversion to Electron Volts

The options provided for the photon's energy are in different units, including electron volts (eV). It is often more convenient to express photon energy in electron volts when dealing with atomic and molecular processes, as Joules represent a very large amount of energy at the quantum scale. To convert energy from Joules to electron volts, we use the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).

\[E (\text{in eV}) = \frac{E (\text{in J})}{1.602 \times 10^{-19} \text{ J/eV}}\]

\[E (\text{in eV}) = \frac{3.3748 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}}\]

\[E (\text{in eV}) \approx 2.1066 \text{ eV}\]

Rounding this value, we get approximately \(2.1 \text{ eV}\).

Therefore, the energy of a photon with a wavelength of \(5890 \text{ Å}\) emitted by a sodium vapor lamp is approximately \(2.1 \text{ eV}\).