Question

In fluorescence, an atom absorbs a photon with a wavelength of 311 nm. This energy radiates out of the atom in two stages. The wavelength of the second emitted photon is 694 nm. What is the wavelength of the second emitted photon?

Answer 1

The energy of the second emitted photon is

$$E=\frac{hc}\lambda=2.86·10^{-19}\text{ J, or 1.78 eV.}$$