Question

The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series. Calculate the wavelengths of the first four members of the Lyman series

Short answer

$$\begin{gathered} \Rightarrow WavelengthsofthefirstfourmembersoftheLymanseries- \\ Forn=2 \\ \lambda =121.57nm \\ Forn=3 \\ \lambda =102.57nm \\ Forn=4 \\ \lambda =97.26nm \\ Forn=5 \\ \lambda =94.98nm \end{gathered}$$

Step-by-step solution

Given Information

Given : The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series.

Formula

$$\begin{gathered} WeknowtheformulaforHydrogenspectrumforanyseries \\ \frac{1}{m^2}-\frac{1}{n^2}=\frac{91.18\left(nm\right)}{\lambda } \\ Wherem=1(Given) \\ n=2,3,4,... \\ So, \\ \frac{1}{1^2}-\frac{1}{n^2}=\frac{91.18\left(nm\right)}{\lambda },Letbetheequation\left(1\right) \end{gathered}$$

Now  we will calculate the wavelengths of the first four members of the Lyman series 

$$\begin{gathered} \Rightarrow Forn=2 \\ Putn=2inequation\left(1\right) \\ \frac{1}{1^2}-\frac{1}{2^2}=\frac{91.18\left(nm\right)}{\lambda } \\ 1-\frac{1}{4}=\frac{91.18\left(nm\right)}{\lambda } \\ \lambda =\frac{91.18\left(nm\right)}{0.75}=121.5733\approx 121.57nm \\ \Rightarrow Forn=3 \\ Putn=3inequation\left(1\right) \\ \frac{1}{1^2}-\frac{1}{3^2}=\frac{91.18\left(nm\right)}{\lambda } \\ \frac{1}{1}-\frac{1}{9}=\frac{91.18\left(nm\right)}{\lambda } \\ \lambda =\frac{91.18\left(nm\right)}{0.8888}=102.57nm \\ \Rightarrow Forn=4 \\ Putn=4inequation1 \\ \frac{1}{1^2}-\frac{1}{4^2}=\frac{91.18\left(nm\right)}{\lambda } \\ \lambda =\frac{91.18\left(nm\right)}{0.9375}=97.2586\approx 97.26nm \\ \Rightarrow Forn=5 \\ Putn=5inequation\left(1\right) \\ \frac{1}{1^2}-\frac{1}{5^2}=\frac{91.18\left(nm\right)}{\lambda } \\ \lambda =\frac{91.18\left(nm\right)}{0.96}=94.9791\approx 94.98nm \end{gathered}$$