91Ó°ÊÓ

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,

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

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

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

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

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