Determine the net energy absorbed by an atom during a process in...

The net energy absorbed by the atom is the difference between the energy of the absorbed photon and the energy of the emitted photon.

$\Delta E = E_{\text{absorbed}} - E_{\text{emitted}}$

$\Delta E = \dfrac{hc}{\lambda_{\text{absorbed}}} - \dfrac{hc}{\lambda_{\text{emitted}}}$

Substituting the given values, $\lambda_{\text{absorbed}} = 500 \text{ nm} and$\lambda_{\text{emitted}} = 700 \text{ nm}$:<br><br>$\Delta E = hc \left( \dfrac{1}{500 \times 10^{-9}} - \dfrac{1}{700 \times 10^{-9}} \right)<br><br>Usinghc = 6.63 \times 10^{-34} \times 3 \times 10^8 = 19.89 \times 10^{-26} \text{ J m}$:<br><br>$\Delta E = 19.89 \times 10^{-26} \times 10^9 \left( \dfrac{1}{500} - \dfrac{1}{700} \right)$<br><br>$\Delta E = 19.89 \times 10^{-17} \left( \dfrac{7 - 5}{3500} \right)<br><br>\Delta E = 19.89 \times 10^{-17} \times $\dfrac{2}{3500}<br><br>$\Delta E = $\dfrac{39.78 \times 10^{-17}}{3500} \approx$1.136 \times 10^{-19} \text{ J}$<br><br>Rounding to two significant figures, the net energy absorbed is$1.1 \times 10^{-19} \text{ J}$.$