Question

The ratio of the shortest wavelength of two spectral series of the hydrogen spectrum is found to be 9. The spectral series is :
(A)- Paschen and Lyman
(B)- Lyman and Paschen
(C)- Brackett and Pfund
(D)- Balmer and Brackett

Answer 1

Hint:. The set of wavelengths arranged in a sequential fashion which characterizes light or any electromagnetic radiation emitted by energized atoms is known as Spectral series. A hydrogen atom has the simplest atomic system found in nature and thus it produces the simplest spectral series.

Complete step by step answer:
-When the beam of light or any radiation is made to enter the device through a slit, each component of the light or radiation form images of the source which can be visualized when resolved under a spectroscope.
-Higher the wavelength, more apart will be the spectral lines. Lower the wavelength, closer the spectral lines.
-The formula which relates to the energy difference between the various levels of Bohr’s model and the wavelengths of absorbed and emitted photons is known as Rydberg formula which is mathematically expressed as-
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{{{n}_{l}}^{2}}-\dfrac{1}{{{n}_{h}}^{2}} \right)$
where $\lambda $ = wavelength;
            R = Rydberg constant $=1.09737\times {{10}^{7}}{{m}^{-1}}$
            Z = atomic number
            ${{n}_{l}}$= lower energy level
            ${{n}_{h}}$= higher energy level

-There are following types of spectral series observed, which are- Lyman; Balmer; Paschen; Brackett; Pfund; Humphreys series.
-Let us focus on the five types of spectral series with context to the question, that is Lyman, Balmer, Paschen, Brackett and Pfund.
-According to Bohr’s atomic model, Lyman series are displaced when electron transition takes place from higher energy states such as ${{n}_{h}}$= 2, 3, 4, 5, 6,….. to ${{n}_{l}}=1$. -According to Bohr’s atomic model, Balmer series is observed when electron transition takes place from higher energy states such as ${{n}_{h}}$= 3, 4, 5, 6, 7,…. to ${{n}_{l}}=2$.
- According to Bohr’s atomic model, Paschen series are displaced when electron transition takes place from higher energy states such as ${{n}_{h}}$= 4, 5, 6,7, ….. to ${{n}_{l}}=3$. From this series, all the subsequent series overlaps.
- According to Bohr’s atomic model, Brackett series are displaced when electron transition takes place from higher energy states such as ${{n}_{h}}$= 5, 6, 7, 8,….. to ${{n}_{l}}=4$.
- According to Bohr’s atomic model, Pfund series are displaced when electron transition takes place from higher energy states such as ${{n}_{h}}$= 6, 7, 8, 9,….. to ${{n}_{l}}=5$.

-From the above knowledge, let us now try to find the spectral series:
$\dfrac{\dfrac{1}{{{\lambda }_{2}}}=R\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right){{Z}^{2}}}{\dfrac{1}{{{\lambda }_{1}}}=R\left( \dfrac{1}{{{m}_{1}}^{2}}-\dfrac{1}{{{m}_{2}}^{2}} \right){{Z}^{2}}}$ where ${{n}_{2}}=\infty ;{{m}_{2}}=\infty $
For the shortest wavelengths, both ${{n}_{2}}$ and ${{m}_{2}}$are in ratio 9. Therefore, the value of ${{n}_{1}}$and ${{m}_{1}}$will be,
$\begin{align}
 & \dfrac{{{\lambda }_{m}}}{{{\lambda }_{n}}}=\dfrac{{{m}_{1}}^{2}}{{{n}_{1}}^{2}}=\dfrac{9}{1} \\
 & \Rightarrow {{m}_{1}}=3;{{n}_{1}}=1 \\
\end{align}$
Therefore the spectral series will be Paschen and Lyman.
So, the correct answer is “Option A”.

Note: You should make a note here that the Rydberg formula is only valid for Hydrogen and Hydrogen like elements only. All the wavelengths of the Lyman series fall under the ultraviolet band. All the wavelengths of the Balmer series are observed in a visible part of the electromagnetic spectrum. All the wavelength of the Paschen series falls under the infrared band. All the wavelength of the Brackett series falls under the infrared region. All the wavelengths of the Pfund series fall under the infrared region.