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.

Lyman series in hydrogen spectrum starts when value of$m=1$

We have to calculate the wavelengths of first four member of Lyman series.

(a)The first member of Lyman series is , $m=1ton=2$

By hydrogen emission spectrum wavelengths is given by,

$\lambda =\frac{91.18}{{\displaystyle \frac{1}{m^2}}-{\displaystyle \frac{1}{n^2}}}\mathit{n}\mathit{m}$

By substituting the value of $m=1$and $n=2$ in above formula , we get

$$\begin{gathered} \lambda =\frac{91.18}{{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}}}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{4}}}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{4-1}{4}}}\mathbf{}\mathit{n}\mathit{m} \\ =\frac{364.72}{3}\mathit{n}\mathit{m} \\ \mathbf{}=121.573\mathbf{nm} \end{gathered}$$

(b) The wavelength of second member of Lyman series is ,$m=1ton=3$

By substituting the value of $m=1$and $n=3$ in above wavelength formula

$$\begin{gathered} \lambda =\frac{91.18}{{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{3^2}}}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{9}}}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{9-1}{9}}}\mathbf{}\mathit{n}\mathit{m} \\ =\frac{820.62}{8}\mathit{n}\mathit{m} \\ =102.577\mathit{n}\mathit{m} \end{gathered}$$

(c) The wavelength of third member of Lyman series is ,$m=1ton=4$

By substituting the value of $m=1$and $n=4$ in above wavelength ,

$$\begin{gathered} \lambda =\frac{91.18}{{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{4^2}}}\mathbf{}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{16}}}\mathbf{}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{16-1}{16}}}\mathbf{}\mathit{n}\mathit{m} \\ =\frac{1458.88}{15}\mathit{n}\mathit{m} \\ =97.258\mathbf{}\mathit{n}\mathit{m} \end{gathered}$$

(d) The wavelength of fourth member of Lyman series is ,$m=1ton=5$

By substituting the value of $m=1$ and $n=5$ in above wavelength ,

$$\begin{gathered} \lambda =\frac{91.18}{{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{5^2}}} \\ =\frac{91.18}{{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{25}}} \\ =\frac{91.18}{{\displaystyle \frac{25-1}{25}}} \\ =\frac{2279.5}{24} \\ =94.979 \end{gathered}$$nm