Question

What is the photon energy for yellow light from a highway sodium lamp at a wavelength of 589 nm?

Short answer

The photon energy of the light is 2.1 eV.

Step-by-step solution

Describe the expression of energy of the photon

The energy of a photon of wavelength $\lambda$is given by,

$E=\frac{hc}{\lambda }$……. (1)

Here, h is Planck’s constant, and c is the speed of light.

Determine the photon energy for yellow light from a highway sodium lamp

Substitute the below values in eq 1.

$$\begin{gathered} \mathrm{\lambda }=589\mathrm{nm} \\ \mathrm{h}=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the photon energy of the light is 2.1 eV.