Energy of an ultraviolet photon

Problem

Calculate the energy of an ultraviolet light photon with wavelength $\lambda = 400\;\text{nm}$. Express the result in eV.

Solution

With the handy constant $hc \approx 1240\;\text{eV·nm}$:

$E = \frac{hc}{\lambda} = \frac{1240\;\text{eV·nm}}{400\;\text{nm}}$

$\ev{E \approx 3{,}1\;\text{eV}}$

This is at the boundary between violet and near ultraviolet: higher energy than an “average” visible photon, consistent with the fact that a shorter $\lambda$ corresponds to a larger $E$.

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Topics: Fisica quantistica Concepts: Fotone