What is the numbar of photons …

Energy of photon (E) = {tex}\frac{\mathrm{hc}}{\lambda}{/tex}
h = 6.626 {tex}\times{/tex} 10^-34 Js, c= 3{tex}\times{/tex}10⁸ ms^-1, {tex}\lambda{/tex} = 4000 pm = 4000 {tex}\times{/tex} 10^-12 = 4 {tex}\times{/tex}10^-9 m
{tex}\therefore{/tex} Energy of photon (E) = {tex}\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(4 \times 10^{-9} \mathrm{m}\right)}{/tex}= 4.969 {tex}\times{/tex} 10^-17 J
Now, 4.969 {tex}\times{/tex} 10^-17 J is the energy of photon = 1
{tex}\therefore{/tex}1 J is the energy of photons = {tex}\frac{1}{4.969 \times 10^{-17}}{/tex}= 2.012 {tex}\times{/tex}10¹⁶ photons.