Question

Calculate the wavelength of 1st and 2nd lines in the Paschen series.

Answer 1

Hint: When electron jumps from higher energy level to the \[{3^{rd}}\] orbit, the spectrum of the hydrogen atom that is observed is called the Paschen series. We know when an electron jumps from higher orbit to a lower orbit it releases some energy in the form of electromagnetic radiation or photons.

Formula used:
The wave number of Paschen series is given by,
\[\dfrac{1}{\lambda } = R(\dfrac{1}{{{3^2}}} - \dfrac{1}{{{n^2}}})\]
where, $\lambda$ is the wavelength of the emitted photon, $R$ is the Rydberg constant and $n$ is the orbit number from which the electron jumps to the \[{3^{rd}}\] orbit.
The value of Rydberg’s constant is \[R = 1.1 \times {10^7}{m^{ - 1}}\].

Complete step by step answer:
In the hydrogen spectrum we find different lines in the spectrum due to the variation of wavelength of the photons that releases from the atom. When an electron jumps from any higher orbit to \[{3^{rd}}\] orbit to the lines that are found in the spectrum is called the Paschen series.

The first line of the Paschen series is the line when an electron jumps from the \[{4^{th}}\] to \[{3^{rd}}\]orbit and the second line of paschen series is the line found when an electron jumps from \[{5^{th}}\]orbit to \[{3^{rd}}\] orbit. Now, the wave number of Paschen series is given by,
\[\dfrac{1}{\lambda } = R(\dfrac{1}{{{3^2}}} - \dfrac{1}{{{n^2}}})\]
So, putting the value for first line \[n = 4\] we have,
\[\dfrac{1}{{{\lambda _1}}} = R(\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}})\]
\[\Rightarrow \dfrac{1}{{{\lambda _1}}} = 1.1 \times {10^7}(\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}})\]
Upon simplifying we have ,
\[{\lambda _1} = \dfrac{{144}}{{7 \times 1.1}} \times {10^{ - 7}}\]
\[\Rightarrow {\lambda _1} = 1870\,nm\]
Now, putting the value for second line \[n = 5\]we have,
\[\dfrac{1}{{{\lambda _2}}} = 1.1 \times {10^7}(\dfrac{1}{{{3^2}}} - \dfrac{1}{{{5^2}}})\]
\[\Rightarrow {\lambda _2} = \dfrac{{225}}{{16 \times 1.1}} \times {10^{ - 7}}\]
\[\therefore {\lambda _2} = 1258\,nm\]

Hence, the wavelength of the first lines of Paschen series is \[1870\,nm\] and the wavelength for second lines is \[1258\,nm\].

Note: The general formula to find the wavelength of the photon is \[\dfrac{1}{\lambda } = R(\dfrac{1}{{{m^2}}} - \dfrac{1}{{{n^2}}})\] where \[m\] is the final orbit of the electron. Depending on the value of \[m\]the hydrogen spectrum is divided into different series. In increasing order of \[m\] the names of the series are Lyman series, Balmer series, Paschen series, Bracket series, Pfund series.