The concept of the photon is introduced by discussion of the process of electromagnetic field quantization within a closed cavity or in an open optical system. A photon is characterized by either a wavelength $\lambda$ or equivalently an energy **E**. Energy of a photon is inversely proportional to the wavelength of a photon. The formula for photon energy is given as,

Where

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Let us discuss the problems related to the photon energy.

### Solved Examples

**Question 1: **Calculate the photon energy if the wavelength is 640nm?

** Solution: **

Given parameters are,

$\lambda$ = 640nm

We know the values of c and h; c = 3$\times10^{8}$m/s and h = 6.626$\times10^{-34}$Js

Photon energy formula is,

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

E = $\frac{6.626\times10^{-34}\times3\times10^{8}}{640\times10^{-9}}$ = 0.031$\times10^{-17}$J

**Question 2: **If the energy of a photon is 350$\times10^{-10}$J, calculate the wavelength of that photon?

** Solution: **

Given parameters are,

E = 350$\times10^{-10}$J

We know the values of c and h; c = 3$\times10^{8}$m/s and h = 6.626$\times10^{-34}$Js

Photon energy formula is,

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

So, $\lambda$ = $\frac{hc}{E}$

$\lambda$ = $\frac{6.626\times10^{-34}\times3\times10^{8}}{350\times10^{-10}}$ = 0.056$\times10^{-16}$m

Given parameters are,

$\lambda$ = 640nm

We know the values of c and h; c = 3$\times10^{8}$m/s and h = 6.626$\times10^{-34}$Js

Photon energy formula is,

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

E = $\frac{6.626\times10^{-34}\times3\times10^{8}}{640\times10^{-9}}$ = 0.031$\times10^{-17}$J

Given parameters are,

E = 350$\times10^{-10}$J

We know the values of c and h; c = 3$\times10^{8}$m/s and h = 6.626$\times10^{-34}$Js

Photon energy formula is,

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

So, $\lambda$ = $\frac{hc}{E}$

$\lambda$ = $\frac{6.626\times10^{-34}\times3\times10^{8}}{350\times10^{-10}}$ = 0.056$\times10^{-16}$m