An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 375 nm and then immediately emits a photon whose associated wavelength is 580 nm . How much net energy is absorbed by the atom in this process?

The absorption of a photon by an atomic electron occurs in the process of the photoelectric effect, in which the photon loses all its energy to the atomic electron, which in turn is released from the atom. This process requires the incident photon to have an energy greater than the binding energy of the orbital electron.

The Energy that is absorbed is equal to the difference in energy of those photons given by,

$\mathbf{∆}\mathbf{E}\mathbf{=}\frac{\mathbf{hc}}{{\mathbf{\lambda }}_{\mathbf{1}}}\mathbf{-}\frac{\mathbf{hc}}{{\mathbf{\lambda }}_{\mathbf{2}}}$ ….. (1)

Here, h is plank constant, c is the speed of the light, ${\mathit{\lambda }}_{\mathbf{1}}$, and ${\mathit{\lambda }}_{\mathbf{2}}$ are the wavelengths.

Substitute $6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$for h, $3\times {10}^8\mathrm{m}/\mathrm{s}$for c , $375\times {10}^{-9}\mathrm{m}$for $\lambda$, and $580\times {10}^{-9}\mathrm{m}$for ${\lambda }_2$in equation (1).

role="math" localid="1661838957695" $$\begin{gathered} ∆E=\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{375\times {10}^{-9}\mathrm{m}}-\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{580\times {10}^{-9}\mathrm{m}} \\ =\frac{\left(19.878\times {10}^{-26}\mathrm{J}.\mathrm{m}\right)}{375\times {10}^{-9}\mathrm{m}}-\frac{\left(19.878\times {10}^{-26}\mathrm{J}.\mathrm{m}\right)}{580\times {10}^{-9}\mathrm{m}} \\ =0.0530\times {10}^{-17}\mathrm{J}-0.0343\times {10}^{-17}\mathrm{J} \\ =1.87\times {10}^{-19}\mathrm{J} \\ ∆E=\left(1.87\times {10}^{-19}\mathrm{J}\right)\left(\frac{6.242\times {10}^{18}\mathrm{eV}}{1\mathrm{J}}\right) \\ =1.17\mathrm{eV} \end{gathered}$$

Hence, the required energy is $1.17\mathrm{eV}$.