`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

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)`