Question

What is the approximate wavelength of the 1st line of paschen series of hydrogen atom?
\[(1){\text{ 1870nm}}\]
\[(2){\text{ 2460nm}}\]
\[(3){\text{ 656nm}}\]
\[(4){\text{ 122nm}}\]

Answer 1

Hint: Paschen series corresponds to the transitions between the higher and lower energy states with principal quantum number 3 or higher. The Paschen series does not lie in the visible spectra. To calculate the first line wavelength in this series, we can use Rydberg's formula.

Complete step-by-step answer:
We know that spectral emission occurs when an electron transitions from a higher energy state to a lower energy state. The lower energy state is designated as n’ and the higher energy state is designated as n. The spectral lines obtained due to the emissions are grouped into a series according to their lower energy state (n’).
The Paschen series lies in the infrared region.
The wavelength of the lines in all the series in the spectrum is described by a simple relation, called the Rydberg’s Formula, which is given as-
\[\dfrac{1}{\lambda } = {R_H}{z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right),{n_2} > {n_1}\]
Where, \[\lambda \]= wavelength
\[{R_H}\]= Rydberg constant,
The value of Rydberg’s constant is \[1.09677 \times {10^7}{m^{ - 1}}\]
The value of \[{n_1}\] is generally fixed for each series. For the Paschen series, \[{n_1} = 3\]
As we know, to calculate the first line wavelength of Paschen series, we have to take \[{n_2} = 4\]
And for hydrogen atom,atomic number will be \[z = 1\]
Substituting all these values in the formula we get,
$\dfrac{1}{\lambda } = 1.09677 \times {10^7} \times {\left( 1 \right)^2} \times \left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right)$
$ \Rightarrow \dfrac{1}{\lambda } = 1.09677 \times {10^7} \times \left( {\dfrac{1}{9} - \dfrac{1}{{16}}} \right)$
$ \Rightarrow \dfrac{1}{\lambda } = 1.09677 \times {10^7} \times \left( {\dfrac{{16 - 9}}{{16 \times 9}}} \right)$
$ \Rightarrow \dfrac{1}{\lambda } = 1.09677 \times {10^7} \times \dfrac{7}{{16 \times 9}}$
$ \Rightarrow \lambda = \dfrac{{16 \times 9}}{{1.09677 \times {{10}^7} \times 7}}$
Doing the calculations, we get,
\[ \Rightarrow \lambda = 18.7 \times {10^{ - 7\,}}m\]
$ \Rightarrow \lambda = 1870\,nm$

Hence, option (A) is the correct answer.

Additional information:
For a hydrogen atom, the spectral lines are given special names. For n’=1 the series is called the Lyman series, which lies in the ultraviolet region. n’=2 is given the name Balmer series which has a few lines in the visible region, n’=3 is the Paschen series lying in the infrared region and there are others for n’=4, 5 and 6 named as Brackett series, Pfund series and Humphrey series respectively.

Note:
It is important to remember the n’ value for the series as it is fixed for every series, for hydrogen atoms. We should not forget to mention the units I.e. metres. The value of Rydberg constant is different for heavier metals so it is important to consider it.