Question

Write the different types of Hydrogen Spectral series. The Lyman series of the Hydrogen spectrum lies in the ultraviolet region. Why?

Answer 1

Hint: The wave numbers of spectral lines in each series can be calculated using Rydberg's equation as follows.
$\dfrac{1}{\lambda } = R{z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ --(i)
Where,
n1and n2 are the principal quantum numbers of orbits corresponding to the electronic transition.
R = Rydberg's constant = 1,09,677 cm-1
Z = atomic number
Using the above formula to get the range of wavelengths of the Lyman series, it must lie in the range of UV wavelength.

Complete step by step answer:
First let us understand what an atomic spectra of hydrogen is.
Hydrogen atomic spectra: When we apply a very high potential to the hydrogen gas at low pressure in a discharge tube, it starts to emit a very bright light. And when it is passed through a prism or grating it gets separated into several lines of radiations and forms a line spectrum. The hydrogen atomic spectrum consists of separate and different spectral lines corresponding to the different wavelengths.
These spectral lines are formed due to electronic transitions of electrons from one energy level to another. These lines can be divided into five series according to their range of wavelengths as follows:
 

Spectral series	Spectral region	n1	n2
1. Lyman series	Ultra-violet	1	2,3,4,5,6,7,_ _ _ _
2. Balmer series	Visible	2	3,4,5,6,7,_ _ _ _
3. Paschen series	near infra-red	3	4,5,6,7,_ _ _ _
4. Brackett series	infra-red	4	5,6,7,_ _ _ _
5. Pfund series	far infra-red	5	6,7,_ _ _ _

The wave numbers of spectral lines in each spectral series can be calculated using Rydberg's equation as follows:
$\dfrac{1}{\lambda } = R{z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ --(i)
Where,
n1and n2 are the principal quantum numbers of orbits corresponding to the electronic transition. 
R = Rydberg's constant = 1,09,677 cm-1 
Z = atomic number

From the above equation,
 For Lyman series:
n1 = 1
and z = 1 (atomic no. of hydrogen is 1)
then equation (i) becomes:
$\dfrac{1}{\lambda } = R{(1)^2}\left[ {\dfrac{1}{{{{\left( 1 \right)}^2}}} - \dfrac{1}{{n_2^2}}} \right] = R\left[ {1 - \dfrac{1}{{n_2^2}}} \right]$ ---(ii)
For calculating longest wavelength putting n2 = 2 in equation (ii) we get:
${\lambda _{\max }}$ = 121.57nm
And for calculating shortest wavelength putting n2 = $\infty $we get
 ${\lambda _{\min }}$ = 91.18nm.
Since, range of ultraviolet region is 10-400nm
Therefore, the entire range of Lyman series lies in the ultraviolet region.

Note:
 Every element has its own characteristic atomic line spectrum. There is a regular trend in the line spectrum of each element. And these spectra can be also considered as the fingerprints of these elements.
Hydrogen, being the smallest atom, has the simplest line spectrum among all the elements. The line spectra become complex as the atomic number increases.