91Ó°ÊÓ

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$\left(\mathrm{n}→∞\right)$

Since m=1,

localid="1651145503134" $$\begin{gathered} {\mathrm{λ}}_{\mathrm{n}→\mathrm{m}}=\frac{{\mathrm{λ}}_0}{\frac{1}{{\mathrm{m}}^2}−\frac{1}{{\mathrm{n}}^2}} \\ {\mathrm{λ}}_{max}=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{1^2}−\frac{1}{2^2}}(\text{substitute}\mathrm{m}=1\text{and}\mathrm{n}=2) \\ {\mathrm{λ}}_{max}=121.57\mathrm{nm} \\ {\mathrm{λ}}_{min}=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{1^2}−0}(\text{substitute}\mathrm{m}=1\text{and}\mathrm{n}→\mathrm{∞}) \\ {\mathrm{λ}}_{min}=91.18\mathrm{nm} \end{gathered}$$

Since, m = 2,

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

Since, m = 3,

localid="1651145573576" $\begin{array}{l}{\mathrm{λ}}_{\max }=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{3^2}−\frac{1}{4^2}}\left(\text{substitute}\mathrm{m}=3\text{and}\mathrm{n}=4\right)\text{​}\\ {\mathrm{λ}}_{\max }=1875.7\text{ }\mathrm{nm}\text{​}\\ {\text{λ}}_{\min }=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{1^3}−0}\left(\text{substitutem}=3\text{andn}→∞\right)\text{​}\\ {\text{λ}}_{\min }=820.6\text{ â¶Ä‰}\mathrm{nm}\text{​}\end{array}$

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:

$\begin{array}{l}{\mathrm{λ}}_{\mathrm{n}→\mathrm{m}}=\frac{{\mathrm{λ}}_0}{\frac{1}{{\mathrm{m}}^2}−\frac{1}{{\mathrm{n}}^2}}\text{​}\\ {\mathrm{λ}}_{3→2}=\frac{91×\left(18×{10}^{−19}\right)}{\frac{1}{2^2}−\frac{1}{3^2}}\left(\text{substitute}\mathrm{m}=2\text{and}\mathrm{n}=3\right)\\ {\mathrm{λ}}_{3→2}=656.5\text{ }\mathrm{nm}\\ {\text{λ}}_{4→2}=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{2^2}−\frac{1}{4^2}}\left(\text{substitutem}=2\text{andn=4}\right)\text{​}\\ {\text{λ}}_{4→2}=486.3\text{ â¶Ä‰}\mathrm{nm}\\ {\text{λ}}_{4→2}=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{2^2}−\frac{1}{5^2}}\left(\text{substitutem}=2\text{andn=5}\right)\text{​}\\ {\text{λ}}_{5→2}=434.2\text{ â¶Ä‰}\mathrm{nm}\text{​}\\ {\text{λ}}_{6→2}=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{2^2}−\frac{1}{6^2}}\left(\text{substitutem}=2\text{andn=6}\right)\text{​}\\ {\text{λ}}_{6→2}=410.3\text{ }\mathrm{nm}\text{​}\\ {\text{λ}}_{7→2}=\frac{91×\left(18×{10}^{−9}\right)}{\frac{1}{2^2}−\frac{1}{5^2}}\left(\text{substitutem}=2\text{andn=7}\right)\text{​}\\ {\text{λ}}_{7→2}=397.4\text{ â¶Ä‰}\mathrm{nm}\text{​}\end{array}$

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