Question

I FIGURE EX38.24 is an energy-level diagram for a simple atom. What wavelengths, in \mathrm{nm}, appear in the atom's (a) emission spectrum and (b) absorption spectrum?

$n=3⟶E_3=4.00\mathrm{eV}$

$n=2-E_2=1.50\mathrm{eV}$

FIGURE EX38.24 n=1-E_{1}=0.00 \mathrm{eV}

Short answer

a)${\lambda }_{21}=828.26\mathrm{nm},{\lambda }_{31}=310.6\mathrm{nm},{\lambda }_{32}=497.36\mathrm{nm}$

b) ${\lambda }_{12}=828.26\mathrm{nm},{\lambda }_{13}=310.6\mathrm{nm}$

Step-by-step solution

part (a) step 1 : given information

We can start the solution with expression for wavelength from transition

$$\begin{gathered} \lambda =\frac{hc}{\Delta E} \\ {\lambda }_{21}=\frac{hc}{E_2-E_1}(transition2\to 1) \\ {\lambda }_{21}=\frac{6.63·{10}^{-34}·3·{10}^8}{1.5·1.6·{10}^{-19}-0}\left(substitute\right) \\ {\lambda }_{21}=828.26\mathrm{nm} \\ {\lambda }_{31}=\frac{hc}{E_3-E_1}(transition3\to 1) \\ {\lambda }_{31}=\frac{6.63·{10}^{-34}·3·{10}^8}{4·1.6·{10}^{-19}-0}\left(substitute\right) \\ {\lambda }_{31}=310.6\mathrm{nm} \\ {\lambda }_{32}=\frac{hc}{E_3-E_2}(transition3\to 2) \\ {\lambda }_{32}=\frac{6.63·{10}^{-34}·3·{10}^8}{4·1.6·{10}^{-19}-1.5·1.6·{10}^{-19}}\left(substitute\right) \\ {\lambda }_{32}=497.36\mathrm{nm} \end{gathered}$$

part (b) step 1 : given information

Atom is in the n=1 state, so we have only absorption from $1\to 2$and $1\to 3$. Expressions are same, so wavelengths are same.

$1\to 2:$ ${\lambda }_{12}=828.26\mathrm{nm}$

$1\to 3:$ ${\lambda }_{13}=310.6\mathrm{nm}$