Photon Energy from Wavelength (E = hc/λ)

This is $E = hf$ rewritten in the unit spectroscopists actually work in. Since $c = f\lambda$, the frequency is $c/\lambda$, and substituting gives $E = hc/\lambda$. Nothing new has been claimed — it is the same physical statement wearing different clothes. What changes is the shape of the dependence: energy is now inversely proportional to wavelength, so halving the wavelength doubles the photon energy, and the short end of the spectrum is where all the energy lives.

The number to carry in your head is $hc \approx 1240$ eV·nm, and it makes most of this arithmetic mental. Divide 1240 by the wavelength in nanometres and read electronvolts: 620 nm red gives 2.0 eV, 400 nm violet gives 3.1 eV, the 254 nm line of a germicidal mercury lamp gives 4.9 eV, and a 0.1 nm X-ray gives 12.4 keV. In SI, a 500 nm photon works out to $6.626 \times 10^{-34} \times 2.998 \times 10^{8} / 500 \times 10^{-9} = 3.97 \times 10^{-19}$ J, which is what this page returns.

The form matters because so many thresholds in nature are quoted as energies. Silicon has a band gap of 1.1 eV, so its cutoff wavelength is $1240/1.1 \approx 1130$ nm: every infrared photon longer than that passes through a silicon cell without producing any current at all, and every photon much shorter wastes its surplus as heat. Those two losses together are most of the Shockley–Queisser limit on what a single-junction cell can ever achieve. The same arithmetic explains the width of the visible band. Photons between roughly 1.6 and 3.1 eV are energetic enough to shift electrons between molecular levels — which is what vision and photosynthesis need — and not energetic enough to break the covalent bonds holding the molecule together. Evolution had a narrow window to work in and used it.

One error dominates all others here: the $\lambda$ must be the vacuum wavelength. Light entering glass keeps its frequency and shortens its wavelength by a factor of $n$. Feed that shortened in-medium wavelength into $E = hc/\lambda$ and you will conclude that a photon gains 50% more energy by entering glass, which is nonsense — energy is conserved at the interface. Either convert back to vacuum wavelength, or sidestep the whole question by using $E = hf$, since frequency is the quantity that does not change. Two smaller cautions: the 1240 constant carries its units with it, so it demands nanometres and returns electronvolts, and the same constant in micrometres is 1.240 eV·µm. And because the relation is reciprocal, equal steps in wavelength are not equal steps in energy — going from 400 to 500 nm costs 0.62 eV while going from 1000 to 1100 nm costs 0.11 eV. That asymmetry is exactly why infrared spectroscopy is done in wavenumbers, which are proportional to energy, rather than in wavelength.