Question

In addition to continuous radiation, fluorescent lamps emit sharp lines in the visible region from a mercury discharge within the tube. Much of this light has a wavelength of 436 nm. What is the energy (in J) of one photon of this light?

Answer 1

Answer:

$4.56 x 10^{-19}J$

Explanation:

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 $6.626 x10^{-34}Js$ 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 =$3 x 10^{8}m/s$) while λ is the wavelength of light.

Wavelength λ = 436nm = $436 x 10^{-9}m$

Therefore the energy E of one photon of this light, using equation (3) is

$E=\frac{6.626 x 10^{-34} x3 x10^{8} }{436 x 10^{-9} } = 4.56 x 10^{-19} J$