Question

# What is the number of photons of light with a wavelength of 4000 pm that provides 1. 1 of energy?

Solution

Energy (E) of a photon ( =mathrm{hc} / lambda )

Where, ( lambda= ) wavelength of light ( =4000 mathrm{pm}= ) ( 4000 times 10-12 m=4 times 10-9 m )

c = velocity of light in vacuum ( =3 times 108 mathrm{m} / mathrm{s} )

( mathrm{h}= ) Planck's constant ( =6.626 times 10-34 mathrm{Js} )

thefore the energy of photon (E)

( =6.626 times 10-34 mathrm{Js} times 3 times 108 mathrm{m} / mathrm{s} / 4 times 10-9 mathrm{m}= )

( 4.965 times 10-16 mathrm{J} )

Now ( 4.965 times 10-16 mathrm{J} ) is the energy of ( =1 ) photon

Therefore ( 1 mathrm{J} ) will be the energy of ( =1 / 4.965 mathrm{x} ) ( 10-16 mathrm{J}=2.014 times 1015 ) photons