In addition to continuous radiation, fluorescent lamps emit some visible lines from mercury. A prominent line has a wavelength of 436 nm. What is the energy (in J) of one photon of it?

Respuesta :

The energy of one photon is 4.56×[tex]10^{-19}[/tex]J

Electromagnetic radiations consist of quanta of energy called photons which have energy, E which is equal to:

E = hν......(1)

where h is the Planck's constant which is  and ν is the frequency of light radiation.

But ν = c/λ ......(2)

Putting equation (2) into (1), we have

E = hc/λ......(3)

c is the speed of light (c =3×[tex]10^{8}[/tex]) while λ is the wavelength of light.

Wavelength λ = 436nm = 3× [tex]10^{-19}[/tex]m

Therefore the energy E of one photon of this light, using equation (3) is (6.626×[tex]10^{-34}[/tex]×3×[tex]10^{8}[/tex])÷4.36×[tex]10^{-9}[/tex] = 4.56×[tex]10^{-19}[/tex]J

To know more about Photon refer to https://brainly.com/question/14649644

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