Question

A gas absorbs a photon of 355 nm and emits it at two wavelengths. If one of the emissions is at 680 nm, what will be the next wavelength emitted?

Answer 1

Conservation of Energy

• According to the Law of conservation energy, the absorbed photon must be equal to the combined energy of the emitted photons
  ${\mathrm{E}}_{\mathrm{A}}={\mathrm{E}}_1+{\mathrm{E}}_2+\dots$
  Here, ${\mathrm{E}}_{\mathrm{A}}$ is the energy of the absorbed photon.
  ${\mathrm{E}}_1\mathrm{and}{\mathrm{E}}_2$are the energy of the emitted photons.

Calculation of the second wavelength

• Step 1: Relationship between energy and the wavelength
  $\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$
  Here, E represents energy
  h represents Planck's constant
  c represents velocity
  $\mathrm{\lambda }$ represents wavelength
• Step 2: Determination of the second wavelength
  The wavelength of the absorbed photon (${\mathrm{\lambda }}_{\mathrm{A}}$) = 355 nm
  The wavelength of the emitted photon (${\mathrm{\lambda }}_1$) = 680 nm
  $$\begin{gathered} \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{A}}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}+\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2} \\ \frac{1}{{\mathrm{\lambda }}_{\mathrm{A}}}=\frac{1}{{\mathrm{\lambda }}_1}+\frac{1}{{\mathrm{\lambda }}_2} \\ \frac{1}{355}=\frac{1}{680}+\frac{1}{{\mathrm{\lambda }}_2} \\ \frac{1}{{\mathrm{\lambda }}_2}=\frac{1}{355}-\frac{1}{680} \\ {\mathrm{\lambda }}_2=743\mathrm{nm} \end{gathered}$$

Therefore, the wavelength of the second emitted photon is 743 nm.