Question

In the following Figure is an energy-level diagram for a simple atom. What wavelengths, in nm, appear in the atom’s (a) emission spectrum and (b) absorption spectrum?

Short answer

(a) The emission spectrum wavelengths are ${\text{λ}}_{21}=828.26\mathrm{nm},{\lambda }_{31}=310.6\mathrm{nm},{\lambda }_{32}=497.36\mathrm{nm}$

(b) The absorption spectrum wavelengths are${\mathrm{\lambda }}_{12}=828.26\mathrm{nm},{\mathrm{\lambda }}_{13}=310.6\mathrm{nm}$

Step-by-step solution

Part (a) Step 1: Given information

The given wavelength from the transition is$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{\Delta E}}$

Part (a) Step 2: Finding the wavelengths of emission spectrum 

The solution can begin with a wavelength equation from the transition.

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{\Delta E}} \\ {\mathrm{\lambda }}_{21}=\frac{\mathrm{hc}}{{\mathrm{E}}_2-{\mathrm{E}}_1}(\mathrm{transition}2\to 1) \\ {\mathrm{\lambda }}_{21}=\frac{6.63\cdot {10}^{-34}\cdot 3\cdot {10}^8}{1.5\cdot 1.6\cdot {10}^{-19}-0}\left(\mathrm{substitute}\right) \\ {\mathrm{\lambda }}_{21}=828.26\mathrm{nm} \\ {\mathrm{\lambda }}_{31}=\frac{\mathrm{hc}}{{\mathrm{E}}_3-{\mathrm{E}}_1}(\text{transition}3\to 1) \\ {\mathrm{\lambda }}_{31}=\frac{6.63\cdot {10}^{-34}\cdot 3\cdot {10}^8}{4\cdot 1.6\cdot {10}^{-19}-0}\left(\mathrm{substitute}\right) \\ {\mathrm{\lambda }}_{31}=310.6\mathrm{nm} \\ {\mathrm{\lambda }}_{32}=\frac{\mathrm{hc}}{{\mathrm{E}}_3-{\mathrm{E}}_2}(\text{transition}3\to 2) \\ {\mathrm{\lambda }}_{32}=\frac{6.63\cdot {10}^{-34}\cdot 3\cdot {10}^8}{4\cdot 1.6\cdot {10}^{-19}-1.5\cdot 1.6\cdot {10}^{-19}}\left(\mathrm{substitute}\right) \\ {\mathrm{\lambda }}_{32}=497.36\mathrm{nm} \end{gathered}$$

Part (b) Step 1: Given information

The chemical make-up of the gas in the stellar atmosphere determines which wavelengths are absorbed.

Part (b) Step 2: Finding the wavelengths of Absorption spectrum

We only have absorption from $1\to 2$and $1\to 3$because the atom is in the n=1 state. Because the expressions are the same, the wavelengths are the same.

$$\begin{gathered} 1\to 2: \\ {\mathrm{\lambda }}_{12}=828.26\mathrm{nm} \\ 1\to 3: \\ {\mathrm{\lambda }}_{13}=310.6\mathrm{nm} \end{gathered}$$