Question

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

Short answer

The wavelengths of visible light in the emission spectrum of the hydrogen atom is${\mathrm{\lambda }}_{3\to 2}=656.5\mathrm{nm},{\mathrm{\lambda }}_{4\to 2}=486.3\mathrm{nm},{\mathrm{\lambda }}_{5\to 2}=434.2\mathrm{nm},{\mathrm{\lambda }}_{6\to 2}=410.3\mathrm{nm}$

Step-by-step solution

Given information

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

Finding wavelengths calculated from the greatest value to lowest value

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

Calculations

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,$397.4\mathrm{nm}<400\mathrm{nm}$