Energy (E) of a photon = hν
Energy `(E_(n))` of ‘n’ photons = nhv
`Rightarrow n=( E_(n)lamda)/(hc)`
Where
`lambda` = wavelength of light = 4000 pm = `4000xx10^(12) m`
c = velocity of light in vacuum = ` 3xx 10^(8)m//s`
h = Planck’s constant = `6.626 xx 10 ^(34)Js`
Substituting the values in the given expression of n:
`n= ((1)xx(4000xx10^(12)))/((6.626xx10^(34)(3 xx 10^(8)))=2.012xx10^(16)`
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are `2.012 xx 10^(16)`
What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy ?
Preshita Prabhakar
Asked: 2 years ago20221031T19:14:00+05:30
20221031T19:14:00+05:30In: General Awareness
what is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy ?
what is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy ?
Leave an answer
Leave an answer
Rupal Uday Mehrotra
Asked: 2 years ago20221029T02:11:00+05:30
20221029T02:11:00+05:30In: Class 11
What is the number of photons of light with a wavelength of 4000 PM that provide 1 joule of energy
What is the number of photons of light with a wavelength of 4000 PM that provide 1 joule of energy
Leave an answer
Leave an answer

0.223×10^35.
2.014 ×10 to the power 15
`lamda = 4000″ pm ” = 4000 xx 10^(12) m = 4xx 10^(9) m`
`E = Nhv = N h(c)/(lamda) :. N = (E xx lamda)/(h xx c) = ((1J) xx (4xx 10^(9) m))/((6.626 xx 10^(34) Js) (3.0 xx 10^(8) ms^(1))) = 2.012 xx 10^(16)` photons