Question

The shortest wavelength for Lyman series is $912\,\overset{\text{o}}{\mathop{\text{A}}}\,$. Find the shortest wavelength for Paschen and Brackett series in the Hydrogen atom.

Answer 1

Hint: When an electron jumps between orbits, it either absorbs or emits energy and radiation corresponding to that energy is given out. The wavelength of the radiation emitted is given by Rydberg's formula. According to it, the wavelength depends on the atomic number and the shell numbers of the orbits. Substituting corresponding values in the relation, the shortest wavelength for certain series of radiation can be calculated.
Formulas used:
$\dfrac{1}{\lambda }=R{{z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$

Complete answer:
When energy is given to an electron in a shell of an atom, it gets excited and hence moves to higher energy levels. When it emits out energy, it moves back to the lower energy level.
When an electron emits energy and comes back to the energy level corresponding to $n=1$ from different higher energy levels, the series of wavelengths emitted are known as Lyman series.
When an electron moves back to $n=2$ from different higher energy levels, the series of wavelengths emitted is known as Balmer series.
When an electron moves back to $n=3$ from different energy levels, the series of wavelengths emitted is known as Paschen series.
When an electron moves back to $n=4$ from different energy levels, the series of wavelengths emitted is known as Brackett series.
The wavelength emitted is calculated according to the Rydberg’s formula is-
$\dfrac{1}{\lambda }=R{{z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$ - (1)
Here, $\lambda $ is the wavelength
$R$ is the Rydberg’s formula
$z$ is the atomic number
${{n}_{1}},\,{{n}_{2}}$ are the number of shells
For Paschen series in Hydrogen, $z=1,\,{{n}_{1}}=3$, for shortest wavelength, ${{n}_{2}}=\infty $
 We substitute given values in eq (1) to get,
$\begin{align}
  & \dfrac{1}{\lambda }=1.1\times {{10}^{7}}{{(1)}^{2}}\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right) \\
 & \Rightarrow \dfrac{1}{\lambda }=1.1\times {{10}^{7}}\left( \dfrac{1}{9} \right) \\
 & \Rightarrow \lambda =\dfrac{9}{1.1\times {{10}^{7}}} \\
 & \Rightarrow \lambda =8182\,\overset{\text{o}}{\mathop{\text{A}}}\, \\
 & \therefore \lambda \approx 8200\,\overset{\text{o}}{\mathop{\text{A}}}\, \\
\end{align}$
Therefore, the shortest wavelength in the Paschen series is $8200\,\overset{\text{o}}{\mathop{\text{A}}}\,$.
For Brackett series in Hydrogen, $z=1,\,{{n}_{1}}=4$, for shortest wavelength, ${{n}_{2}}=\infty $
We substitute given values in the above equation to get,
$\begin{align}
  & \dfrac{1}{\lambda }=1.1\times {{10}^{7}}{{(1)}^{2}}\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right) \\
 & \Rightarrow \dfrac{1}{\lambda }=1.1\times {{10}^{7}}\left( \dfrac{1}{16} \right) \\
 & \Rightarrow \lambda =\dfrac{16}{1.1\times {{10}^{7}}} \\
 & \Rightarrow \lambda =14545.5\,\overset{\text{o}}{\mathop{\text{A}}}\, \\
 & \therefore \lambda \approx 14550\,\overset{\text{o}}{\mathop{\text{A}}}\, \\
\end{align}$
Therefore, the shortest wavelength in the Brackett series is $14550\,\overset{\text{o}}{\mathop{\text{A}}}\,$.
Therefore, the shortest wavelength in Paschen series is $8200\,\overset{\text{o}}{\mathop{\text{A}}}\,$ and the shortest wavelength in Brackett series is $14550\,\overset{\text{o}}{\mathop{\text{A}}}\,$.

Note:
When an electron moves from a higher orbit to lower orbit, energy is given out but when it moves from the lower orbit to higher orbit, it absorbs energy. In Bohr’s orbits, energy is not emitted by the electrons and is in its most stable state. The energy required to move an electron from its orbit to infinity is called excitation energy.