What is the numbar of photons …
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Preeti Dabral 1 year, 5 months ago
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.
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