Question

What is the relationship between wavelength, frequency, and energy of a photon?

Answer 1

Hint: A photon is the smallest discrete amount of electromagnetic light. Photons are individual energy packets of light, also known as quanta. The energy associated with a photon is related to the wavelength and frequency of radiation:
\[\text{E}=\text{h}\nu =\dfrac{\text{hc}}{\lambda }\]

Complete answer:
A photon is the elementary particle of light. It is the smallest discrete energy packet of electromagnetic radiation. Photons can be absorbed or emitted by atoms and molecules. When a photon is absorbed, its entire energy is transferred to that atom or molecule. And when an atom or molecule loses energy, it emits a photon that carries energy exactly equal in amount to the loss in energy of the atom or molecule. Planck established a relation between this change in energy and the frequency of the photon emitted or absorbed in the form of a mathematical equation as given below:
\[\text{E}=\text{h}\nu \]
According to this relation, the energy of a photon is directly proportional to the frequency of light emitted or absorbed. The proportionality constant “h” is known as Planck’s constant and its numerical value is given as:
\[\text{h}=6.63\times {{10}^{-34}}\text{ J s}\]
The wavelength and frequency are interrelated to each other. Wavelength is inversely proportional to the frequency and the relation is given as:
\[\nu =\dfrac{\text{c}}{\lambda }\]
Where c is the speed of light and it is equal to $3\times {{10}^{8}}\text{ m }{{\text{s}}^{-1}}$.
Thus, the energy of a photon can be related to frequency and wavelength by resulting equation of the above two relations:
\[\text{E}=\text{h}\nu =\dfrac{\text{hc}}{\lambda }\]
Hence, this is the relationship between wavelength, frequency, and energy of a photon.

Note:
As the wavelength of light gets shorter, the energy of the photon increases. The energy of a mole of photons that have the wavelength $\lambda $ can be found by multiplying the above equation by Avogadro's number:
\[~{{\text{E}}_{\text{m}}}={{\text{N}}_{\text{A}}}\text{h}\nu =\dfrac{{{\text{N}}_{\text{A}}}\text{hc}}{\lambda }\]