What is the energy of a photon from the sodium D line at 589nm? Convert the energy to electron volts?

The energy of a photon can be calculated using the formula:

E = hc/λ

Where:
E = energy of the photon
h = Planck's constant (6.626 x 10^-34 J s)
c = speed of light (3 x 10^8 m/s)
λ = wavelength of the photon (589 nm = 589 x 10^-9 m)

Plugging in the values:

E = (6.626 x 10^-34 J s * 3 x 10^8 m/s) / (589 x 10^-9 m)
E = 3.37 x 10^-19 J

To convert this energy to electron volts, we can use the conversion factor 1 eV = 1.602 x 10^-19 J:

E = (3.37 x 10^-19 J) / (1.602 x 10^-19 J/eV)
E ≈ 2.11 eV

Therefore, the energy of a photon from the sodium D line at 589 nm is approximately 2.11 electron volts.

The energy of a photon is given by:

$$E = hf = \frac{hc}{\lambda}$$

where:

- $E$ is the energy of the photon (in Joules)
- $h$ is Planck's constant ($6.63\times 10^{-34}$ J s)
- $f$ is the frequency of the photon (in Hz)
- $c$ is the speed of light ($3\times 10^8$ m/s)
- $\lambda$ is the wavelength of the photon (in meters)

For the sodium D line at 589nm, the wavelength is:

$$\lambda = 589 \text{nm} = 589 \times 10^{-9} \text{m}$$

The frequency of the photon is:

$$f = \frac{c}{\lambda} = \frac{3\times 10^8 \text{m/s}}{589 \times 10^{-9} \text{m}} = 5.09\times 10^{14} \text{Hz}$$

The energy of the photon is:

$$E = hf = (6.63\times 10^{-34} \text{J s})(5.09\times 10^{14} \text{Hz}) = 3.37\times 10^{-19} \text{J}$$

To convert the energy to electron volts (eV), we use the conversion factor:

$$1 \text{eV} = 1.602\times 10^{-19} \text{J}$$

Therefore, the energy of the photon in electron volts is:

$$E = \frac{3.37\times 10^{-19} \text{J}}{1.602\times 10^{-19} \text{J/eV}} = \boxed{2.11 \text{eV}}$$