Chapter 41 - Atomic Physics - Exercises and Problems - Page 1246: 41

See the detailed answer below.

$$\color{blue}{\bf [a]}$$ To calculate the wavelength, we need to use the formula of $$ \Delta E =E_{\rm emitted}= \frac{hc}{\lambda} $$ So, $$ \lambda = \frac{hc}{\Delta E} $$ For each transition, the energy difference \( \Delta E \) is given by: $$ \Delta E = E_{\text{higher}} - E_{\text{lower}} $$ So $$ \lambda = \frac{hc}{(E_{\text{higher}} - E_{\text{lower}})e} $$ And to get it in nanometers, $$ \lambda = \frac{hc}{(E_{\text{higher}} - E_{\text{lower}})e(10^{-9})} $$ Plug the known $$ \boxed{\lambda = \frac{(6.63\times 10^{-34})(3\times 10^8)}{(E_{\text{higher}} - E_{\text{lower}})(1.6\times 10^{-19})(10^{-9})}} $$ Assuming that the allowed transitions must satisfy the selection rule $ \Delta l = \pm 1 $, the angular momentum quantum number must change by 1 during the transition. $$\color{blue}{\bf [b]}$$ To Identify the spectral region (infrared, visible, or ultraviolet), we need to recall that - Infrared (IR) at: $\lambda > 700 \, \text{nm} $ - Visible at: $ 400 \, \text{nm}\leq \lambda \leq 700 \, \text{nm} $ - Ultraviolet (UV) at: $ \lambda< 400 \, \text{nm} $ $$\color{blue}{\bf [c]}$$ The absorption spectrum only contains transitions that end in the ground $ 2s $ state, so only the $ 2p \to 2s $, $ 3p \to 2s $, and $ 4s \to 2p $ transitions are included in the absorption spectrum. Now, let’s calculate for each transition. See the table below. \begin{array}{|c|c|c|c|} \hline \text{Transition} &\color{blue}{\bf [a]} \;\lambda\;{\rm (nm)} & \color{blue}{\bf [b]}\text{ Type} &\color{blue}{\bf [c]} \text{Absorption} \\ \hline 4s \rightarrow 3p & 2437 & \text{IR} & \text{No}\\\hline 4s \rightarrow 2p & 499 & \text{Visible } & \text{No} \\\hline 3d \rightarrow 3p & 24862 & \text{IR} & \text{No} \\\hline 3d \rightarrow 2p & 612 & \text{Visible } & \text{No} \\\hline 3p \rightarrow 3s & 2702 & \text{IR} & \text{No} \\\hline 3p \rightarrow 2s & 325 & \text{UV} & \text{Yes} \\\hline 3s \rightarrow 2p & 818 & \text{IR} & \text{No} \\\hline 2p \rightarrow 2s & 672 & \text{Visible } & \text{Yes} \\\hline \hline \end{array}