Question

Calculate the shortest and longest wavelength of the Balmer series of ${{\text{H}}_2}$ spectrum. Given ${\text{R = 1}}{\text{.097}} \times {\text{1}}{{\text{0}}^7}{{\text{m}}^{ - 1}}$

Answer 1

Hint:The Hydrogen spectrum includes Lyman, Balmer, Paschen, Brackett, Pfund, and Humphreys series. Shortest wavelength in any series denotes that the energy should be maximum. Longest wavelength denotes that the energy should be minimum. Therefore the transition occurs from the next line of series. The Rydberg formula is applied here.

Complete step by step answer:

There are two types- emission and absorption spectrum. Atomic transitions form a series of lines. This is caused by emission or absorption.
In each series, the wavelength can be calculated by Rydberg formula
\[\dfrac{1}{\lambda } = {{\text{Z}}^2}\dfrac{1}{{{{\text{R}}_{\text{H}}}}}\left( {\dfrac{1}{{{{\text{n}}_1}^2}} - \left. {\dfrac{1}{{{{\text{n}}_2}^2}}} \right)} \right.\]
\[\lambda \] is the wavelength
\[{{\text{R}}_{\text{H}}}\] is Rydberg constant = \[1.097 \times {10^7}{{\text{m}}^{ - 1}}\] for hydrogen atom.
\[{{\text{n}}_1}\] is the principal quantum number of lower energy levels.
\[{{\text{n}}_2}\] is the principal quantum number of higher energy levels.
\[{\text{Z}}\] is the atomic number.
\[\dfrac{1}{\lambda } = {{\text{Z}}^2}\dfrac{1}{{{{\text{R}}_{\text{H}}}}}\left( {\dfrac{1}{{{{\text{n}}_1}^2}} - \left. {\dfrac{1}{{{{\text{n}}_2}^2}}} \right)} \right.\]
Here for short wavelength of hydrogen atom in Balmer series, \[{{\text{n}}_1} = 2\], \[{{\text{n}}_2} = \infty \], ${\text{Z}} = 1$
This is because the shortest wavelength means the energy has to be maximum since energy and wavelength are inversely proportional.
Substituting all these values in the Rydberg formula, we get
\[\dfrac{1}{{{\lambda _{\text{s}}}}} = {\text{R}} \times {1^2}\left( {\dfrac{1}{{{2^2}}} - \left. {\dfrac{1}{{{\infty ^2}}}} \right)} \right.\]
On simplification,
  $\dfrac{1}{\infty } \to 0 \Leftrightarrow \dfrac{1}{{{\lambda _{\text{s}}}}} = {\text{R}} \times 1\left( {\dfrac{1}{4}} \right) \\
  \dfrac{1}{{{\lambda _{\text{s}}}}} = \dfrac{{\text{R}}}{4} \\ $
Taking reciprocal, we get
${\lambda _{\text{s}}} = \dfrac{4}{{\text{R}}}$
Substituting the value of ${\text{R}}$,
${\lambda _{\text{s}}} = \dfrac{4}{{1.097 \times {{10}^7}{{\text{m}}^{ - 1}}}} = 364.6 \times {10^{ - 9}}{\text{m = }}364.6{\text{nm}}$
Now we can calculate the longest wavelength of hydrogen atom in the Balmer series.
 i.e. ${{\text{n}}_1} = 2$, \[{{\text{n}}_2} = 3\], \[{\text{Z}} = 1\]
Longest wavelength means the energy has to be minimum.
Therefore for \[{{\text{H}}_2}\] in Balmer series, Rydberg formula can be written as:
\[\dfrac{1}{{{\lambda _{\text{l}}}}} = {\text{R}} \times {1^2}\left( {\dfrac{1}{{{2^2}}} - \left. {\dfrac{1}{{{3^2}}}} \right)} \right.\]
\[\dfrac{1}{{{\lambda _{\text{l}}}}} = {\text{R}}\left( {\dfrac{1}{4} - \left. {\dfrac{1}{9}} \right)} \right.\]
Taking LCM,
\[\dfrac{1}{{{\lambda _{\text{l}}}}} = {\text{R}}\left( {\dfrac{{9 - 4}}{{9 \times 4}}} \right)\]
On simplifying and solving, we get
\[\dfrac{1}{{{\lambda _{\text{l}}}}} = {\text{R}}\left( {\dfrac{5}{{36}}} \right)\]
$\dfrac{1}{{{\lambda _{\text{l}}}}} = \dfrac{{{\text{5R}}}}{{36}} \Leftrightarrow {\lambda _{\text{l}}} = \dfrac{{36}}{{{\text{5R}}}}$
Substituting the value of ${\text{R}}$, we get
${\lambda _{\text{l}}} = \dfrac{{36}}{{5\left( {1.097 \times {{10}^7}} \right)}} = \dfrac{{36}}{{5.48 \times {{10}^7}}} = 6.56 \times {10^7}{{\text{m}}^{ - 1}} = 656{\text{nm}}$
Hence we can say that shortest wavelength in Balmer series, ${\lambda _s} = 364.6{\text{nm}}$ and longest wavelength in Balmer series, ${\lambda _{\text{l}}} = 656{\text{nm}}$
Note:
Spectral lines are formed by the electronic transitions occurring between different energy levels. All the lines in Lyman series are in UV region, Balmer series are in visible region and the other four are in infrared. Electronic transition forms photons having the same energy. The energy difference between each state is fixed.