Question

The shortest wavelength for Lyman series is $ 912 \mathop A\limits^ \circ $. Find the shortest wavelength for the Paschen and Brackett series in hydrogen.

Answer 1

Hint: The Lyman series is obtained when transition is from any higher orbit to the first orbit. Paschen and Brackett series are obtained when transition is from any higher orbit to the third and fourth orbit respectively. The wavelength is found using Bohr’s formula of spectral lines.
 $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n^2}}} - \dfrac{1}{{{m^2}}}} \right) $

Complete step by step answer:
For the Lyman series, $ n = 1 $. For shortest wavelength in Lyman series, $ m = \infty $
 $
 \Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) \\
 \Rightarrow \dfrac{1}{\lambda } = R \\
 \Rightarrow \lambda = \dfrac{1}{R} \\
 $
Here $ R $ is Rydberg's constant.
Given $ \lambda = 912 \mathop A\limits^ \circ $ ,
 $ \Rightarrow \dfrac{1}{R} = 912 \mathop A\limits^ \circ $
For Paschen series, $ n = 3 $ . For shortest wavelength in Paschen series, $ m = \infty $
 $
 \Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) \\
 \Rightarrow \dfrac{1}{\lambda } = \dfrac{R}{9} \\
 \Rightarrow \lambda = \dfrac{9}{R} \\
 $
 $
 \Rightarrow \lambda = 9\left( {\dfrac{1}{R}} \right) \\
 \Rightarrow \lambda = \left( 9 \right)\left( {912} \right) \\
 \Rightarrow \lambda = 8208\mathop A\limits^ \circ \\
 $
Therefore, the shortest wavelength for the Paschen series is $ 8208\mathop A\limits^0 $ .
For Brackett series, $ n = 4 $ . For shortest wavelength in Brackett series, $ m = \infty $
 $
 \Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) \\
 \Rightarrow \dfrac{1}{\lambda } = \dfrac{R}{{16}} \\
 \Rightarrow \lambda = \dfrac{{16}}{R} \\
 $
 $
 \Rightarrow \lambda = 16\left( {\dfrac{1}{R}} \right) \\
 \Rightarrow \lambda = \left( {16} \right)\left( {912} \right) \\
 \Rightarrow \lambda = 14592\mathop A\limits^ \circ \\
 $
Therefore, the shortest wavelength for the Brackett series is $ 14592\mathop A\limits^ \circ $.
Hence, shortest wavelengths for Paschen and Brackett series in hydrogen are $ 8208\mathop A\limits^ \circ $ and $ 14592\mathop A\limits^ \circ $ respectively.

Additional Information:
There are series of lines observed in the spectrum of hydrogen atoms. These lines indicate different transitions. Series of lines are obtained by different transitions. The obtained series are Lyman, Balmer, Paschen, Brackett and Pfund series. For Balmer series, $ n = 2 $, meaning that the transition takes place from any higher orbit to second orbit. For Pfund series, $ n = 5 $, meaning that the transition takes place from any higher orbit to fifth orbit.

Note:
For the shortest wavelength, the transition takes place from infinity to lower orbit. The shortest wavelength is known as series limit. Always remember the values of the principal quantum number $ n $ for the series obtained by various transitions.