Problem — Energy of a visible photon

Problem

A photon of green light has wavelength $\lambda = 500\;\text{nm}$. Calculate its energy, first in joules and then in electronvolts, using the relation $E = hc/\lambda$.

Solution

Setting up. The energy of a photon depends on its wavelength according to:

$E = \frac{hc}{\lambda}$

with $h = 6{,}63\cdot 10^{-34}\;\text{J·s}$, $c = 3\cdot 10^8\;\text{m/s}$ and $\lambda = 500\;\text{nm} = 500\cdot 10^{-9}\;\text{m} = 5{,}00\cdot 10^{-7}\;\text{m}$.

Calculation in joules.

$E = \frac{(6{,}63\cdot 10^{-34})\cdot(3\cdot 10^8)}{500\cdot 10^{-9}} = \frac{1{,}99\cdot 10^{-25}}{5{,}00\cdot 10^{-7}} \approx 3{,}98\cdot 10^{-19}\;\text{J}$

Conversion to electronvolts ($1\;\text{eV} = 1{,}6\cdot 10^{-19}\;\text{J}$):

$E = \frac{3{,}98\cdot 10^{-19}}{1{,}6\cdot 10^{-19}} \approx 2{,}49\;\text{eV}$

$\ev{E \approx 3{,}98\cdot 10^{-19}\;\mathrm{J} \approx 2{,}49\;\mathrm{eV}}$

A result of a few electronvolts is typical of visible-light photons: these are precisely the energies corresponding to electronic transitions in atoms and molecules responsible for colours.

Connections

Topics: Electromagnetic waves Concepts: Photon Skills: Unit conversion