Using the Rydberg formula,calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

The Rydberg formula is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R_H \approx 1.097 \times 10^7 \, m^{-1}$.
For the Lyman series,the final energy level is $n_f = 1$. The first four spectral lines correspond to transitions from $n_i = 2, 3, 4, 5$ to $n_f = 1$.
Using $\lambda = \frac{1}{R_H \left( 1 - \frac{1}{n_i^2} \right)} = \frac{n_i^2}{R_H (n_i^2 - 1)}$:
$1$. For $n_i = 2$: $\lambda_{21} = \frac{4}{1.097 \times 10^7 (3)} \approx 1.216 \times 10^{-7} \, m = 1216 \, \mathring{A}$.
$2$. For $n_i = 3$: $\lambda_{31} = \frac{9}{1.097 \times 10^7 (8)} \approx 1.026 \times 10^{-7} \, m = 1026 \, \mathring{A}$.
$3$. For $n_i = 4$: $\lambda_{41} = \frac{16}{1.097 \times 10^7 (15)} \approx 0.972 \times 10^{-7} \, m = 972 \, \mathring{A}$.
$4$. For $n_i = 5$: $\lambda_{51} = \frac{25}{1.097 \times 10^7 (24)} \approx 0.950 \times 10^{-7} \, m = 950 \, \mathring{A}$.