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Find the shortest wavelength in Paschen series if, the longest wavelength in Balmer series is 6563 ${A^ \circ }$

Last updated date: 16th Sep 2024
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Hint To answer this question we should know that the Paschen series are defined as the series of the lines in the spectrum of the hydrogen atom which corresponds to the transitions between the state along with the principal quantum number n = 3 and successive higher states. Based on this concept we have to answer this question.

It is given that, Longest wavelength of Balmer series = 6563 ${A^ \circ }$
To find the Shortest wavelength of the Paschen series we have to follow the steps.
We know that,
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n^2}}} - \dfrac{1}{{{m^2}}}} \right)$
For Balmer series
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)$
$\Rightarrow \dfrac{1}{{6563}} = R{Z^2}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right)$
$\Rightarrow \dfrac{1}{{6563}} = R{Z^2}\left( {\dfrac{{9 - 4}}{{36}}} \right)$
$\Rightarrow R{Z^2} = \left( {\dfrac{{36}}{{5 \times 6563}}} \right)\;\;\;...(i)$
For Paschen series
Shortest wavelength when $(m = \infty )$
$\dfrac{1}{{\lambda min}}$=$R{Z^2}\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{\infty }} \right)$
$\dfrac{1}{{\lambda min}} = \dfrac{{36}}{{5 \times 6563}} \times \dfrac{1}{9}$
$\lambda min = 8203.75{A^ \circ }$

Hence, the answer is $8203.75{A^ \circ }$.

Note We should know that the Blamer series is the name which is given to the series of the spectral emission lines of the hydrogen atom and comes as a result of the electron transitions from the higher level down to the energy level with the principal quantum number is 2.
The Balmer series is specifically used in astronomy because the Balmer lines appear in numerous stellar objects due to the abundance of hydrogen in the universe, and therefore they are commonly seen and relatively strong compared to the lines from other elements.