Question

Find the longest wavelength in the Paschen’s series (Given $R=1.097\times {{10}^{7}}{{m}^{-1}}$ )

Answer 1

Hint: For the hydrogen spectrum series, use the Rydberg formula as the expression for the wavelength. Paschen’s series is defined for \[{{n}_{1}}=3\], substitute its value and the appropriate value of \[{{n}_{2}}\]to calculate the longest wavelength.

Formula used: The Rydberg formula:
\[\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)\]

Complete step by step answer:
The wavelengths of the hydrogen spectrum could be calculated by the following formula known as the Rydberg formula:

\[\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)\]

Where, \[\lambda \] is the wavelength of the light emitted in units of \[c{{m}^{-1}}\], R is the Rydberg constant for hydrogen (109,677.581\[c{{m}^{-1}}\]) and \[{{n}_{1}}\], \[{{n}_{2}}\] are integers representing the number of orbital such that \[{{n}_{1}}<{{n}_{2}}\].

The Paschen’s series corresponds to the transitions between the states with principal quantum numbers \[{{n}_{1}}=3\] and \[{{n}_{2}}=4,5,6...\infty \]

In the Rydberg formula, the value of \[{{n}_{1}}=3\], we have to find a \[{{n}_{2}}\] such that the wavelength is the longest possible, as R is a constant.

At \[{{n}_{2}}=4\] the longest wavelength occurs. Substituting all the values in the Rydberg formula, we get the value of the wavelength.

 \[\begin{align}
& \dfrac{1}{\lambda }=R\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)=R\dfrac{7}{144}=109,677.581\dfrac{7}{144}c{{m}^{-1}} \\
 & \Rightarrow \lambda =\dfrac{144}{109,677.581\times 7}cm=1875\times {{10}^{-10}}m \\
 & \text{Also, 1 Angstrom}={{10}^{-10}}\text{meter} \\
\end{align}\]

Therefore, the longest wavelength that can be achieved in Paschen’s series is 1875 angstroms.

Additional Information:
In 1914, Niels Bohr proposed a theory of hydrogen. The excitement in an atom (in this case Hydrogen, with one electron only) is determined by the position of its electron. The higher the number of orbits the electron resides in; more is its energy. When the electron falls to an orbital lower than previously, it loses some amount of energy. This energy is emitted in the form of light with a certain wavelength. This is known as the Hydrogen spectrum.

Note: Since, the reciprocal of wavelength is being calculated in the Rydberg formula, for the longest wavelength, the smallest value of the following expression is desirable.

\[R\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{n}_{2}}^{2}} \right)\].

We can take \[{{n}_{2}}=4,5,6...\infty \], but when \[{{n}_{2}}=4\], that is the maximum possible, the expression takes the least value.