Problem Calculate the energy of a blue visible-light photon (λ=400 nm\lambda = 400\;\text{nm}λ=400nm). Solution In SI units: E=hcλ=6,63⋅10−34⋅3,0⋅108400⋅10−9=1,989⋅10−254,0⋅10−7E = \frac{hc}{\lambda} = \frac{6{,}63\cdot 10^{-34}\cdot 3{,}0\cdot 10^8}{400\cdot 10^{-9}} = \frac{1{,}989\cdot 10^{-25}}{4{,}0\cdot 10^{-7}}E=λhc=400⋅10−96,63⋅10−34⋅3,0⋅108=4,0⋅10−71,989⋅10−25 E≈4,97⋅10−19 J=3,10 eV\ev{E \approx 4{,}97\cdot 10^{-19}\;\text{J} = 3{,}10\;\text{eV}}E≈4,97⋅10−19J=3,10eV (Check: E=1240/400=3,10 eVE = 1240/400 = 3{,}10\;\text{eV}E=1240/400=3,10eV.) This is among the most energetic visible photons, as expected for blue light at the border of the ultraviolet. Links Topics: Quantum physics Concepts: Photon