Question

Calculate the wavelength of the first line in the Lyman series of the hydrogen spectrum (R = 109677 cm-1) . How to do this?

Answer 1

Hint :The Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines when an electron moves from n 2 to n = 1 (where n is the primary quantum number), the electron's lowest energy state, in physics and chemistry. The transitions are named after Greek letters as Lyman-alpha is the transition from n = 2 to n = 1.
 ${\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}\right)$
R = Rydberg’s constant (Also written is RH)
Z = atomic number.

Complete Step By Step Answer:
Let us see what is wave number:
The number of complete wave cycles of an electromagnetic field (EM field) that exist in one metre (1 m) of linear space is referred to as the wave number. In reciprocal metres, the wave number is indicated ( $m^{-1}$ ).
For the Lyman series, n1=1.
For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., $n_2=\mathrm{\infty }.$
 ${\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}\right)$
Substitute the values:
since the electron is de-exited from $1^{st}$ excited state (i.e n=2) to ground state (i.e n=1) for the first line of the Lyman series.
Or, ${\displaystyle \frac{1}{\lambda }}=R\left({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{2^2}}\right)$
Or, ${\displaystyle \frac{1}{\lambda }}=R\left({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{4}}\right)$
Or, ${\displaystyle \frac{1}{\lambda }}=R\left({\displaystyle \frac{4-1}{4}}\right)$
Or, ${\displaystyle \frac{1}{\lambda }}=R\left({\displaystyle \frac{3}{4}}\right)$
Or, $\lambda ={\displaystyle \frac{4}{3R}}$
Or, $\lambda ={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{1}{R}}$
Or, $\lambda ={\displaystyle \frac{4}{3}}\times 912$
Or, $\lambda =121.6nm$
Hence the wavelength of the first line in the Lyman series of the hydrogen spectrum is $\lambda =121.6nm$ .

Note :
We can denote the answer in terms of angstroms but always try to answer the wavelength in terms of nanometers. The atomic hydrogen emission spectrum has been separated into a variety of spectral series, with wavelengths calculated using the Rydberg formula. These spectral lines are caused by electrons in an atom transitioning between two energy levels.