Calculate all the wavelengths of visible light in the emission spectrum of the hydrogen atom.

In astronomy, the spectral series are used to detect hydrogen and calculate red shifts.

We can begin by identifying wavelengths from the greatest value (n=m+1) to the smallest value$\left(\mathrm{n}\to \infty \right)$

Since m=1,

localid="1651145503134" $$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{n}\to \mathrm{m}}=\frac{{\mathrm{\lambda }}_0}{\frac{1}{{\mathrm{m}}^2}-\frac{1}{{\mathrm{n}}^2}} \\ {\mathrm{\lambda }}_{max}=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{1^2}-\frac{1}{2^2}}(\text{substitute}\mathrm{m}=1\text{and}\mathrm{n}=2) \\ {\mathrm{\lambda }}_{max}=121.57\mathrm{nm} \\ {\mathrm{\lambda }}_{min}=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{1^2}-0}(\text{substitute}\mathrm{m}=1\text{and}\mathrm{n}\to \mathrm{\infty }) \\ {\mathrm{\lambda }}_{min}=91.18\mathrm{nm} \end{gathered}$$

Since, m = 2,

localid="1651145549014" $\begin{array}{l}{\mathrm{\lambda }}_{\max }=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{3^2}}\left(\text{substitute}\mathrm{m}=2\text{and}\mathrm{n}=3\right)\text{}\\ {\mathrm{\lambda }}_{\max }=656.5\mathrm{nm}\text{}\\ {\text{λ}}_{\min }=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{1^2}-0}\left(\text{substitutem}=2\text{andn}\to \infty \right)\text{}\\ {\text{λ}}_{\min }=364.72\mathrm{nm}\text{}\end{array}$

Since, m = 3,

localid="1651145573576" $\begin{array}{l}{\mathrm{\lambda }}_{\max }=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{3^2}-\frac{1}{4^2}}\left(\text{substitute}\mathrm{m}=3\text{and}\mathrm{n}=4\right)\text{}\\ {\mathrm{\lambda }}_{\max }=1875.7\mathrm{nm}\text{}\\ {\text{λ}}_{\min }=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{1^3}-0}\left(\text{substitutem}=3\text{andn}\to \infty \right)\text{}\\ {\text{λ}}_{\min }=820.6\mathrm{nm}\text{}\end{array}$

We can observe that visible wavelengths only occur for m=2(n=3) from the results. Starting with n=3, we can now determine visible wavelengths in the hydrogen spectrum:

$\begin{array}{l}{\mathrm{\lambda }}_{\mathrm{n}\to \mathrm{m}}=\frac{{\mathrm{\lambda }}_0}{\frac{1}{{\mathrm{m}}^2}-\frac{1}{{\mathrm{n}}^2}}\text{}\\ {\mathrm{\lambda }}_{3\to 2}=\frac{91\times \left(18\times {10}^{-19}\right)}{\frac{1}{2^2}-\frac{1}{3^2}}\left(\text{substitute}\mathrm{m}=2\text{and}\mathrm{n}=3\right)\\ {\mathrm{\lambda }}_{3\to 2}=656.5\mathrm{nm}\\ {\text{λ}}_{4\to 2}=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{4^2}}\left(\text{substitutem}=2\text{andn=4}\right)\text{}\\ {\text{λ}}_{4\to 2}=486.3\mathrm{nm}\\ {\text{λ}}_{4\to 2}=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{5^2}}\left(\text{substitutem}=2\text{andn=5}\right)\text{}\\ {\text{λ}}_{5\to 2}=434.2\mathrm{nm}\text{}\\ {\text{λ}}_{6\to 2}=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{6^2}}\left(\text{substitutem}=2\text{andn=6}\right)\text{}\\ {\text{λ}}_{6\to 2}=410.3\mathrm{nm}\text{}\\ {\text{λ}}_{7\to 2}=\frac{91\times \left(18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{5^2}}\left(\text{substitutem}=2\text{andn=7}\right)\text{}\\ {\text{λ}}_{7\to 2}=397.4\mathrm{nm}\text{}\end{array}$

Because only the first four wavelengths are visible,$397.4\mathrm{nm}<400\mathrm{nm}$