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Emission transitions in the Paschen series end at orbit n=3 and start from orbit n and can be represented as \[v=3.29X{{10}^{15}}(Hz)[1/{{3}^{2}}-1/{{n}^{2}}].\] Calculate the value of n if the transition is observed at 1285nm. Find the region of the spectrum.

Answer Verified Verified
Hint: The emission spectrum of hydrogen decreases in a regular way as sets called spectral series. The series of spectra for hydrogen were subsequently discovered as Lyman, Paschen, Brackett, and Pfund series. The wavelength of the Paschen series limit is 820.1 nm.

Complete answer:
The given frequency of the emission transition in the Paschen series
\[v=3.29X{{10}^{15}}(Hz)[1/{{3}^{2}}-1/{{n}^{2}}].\] -- (1)
And also, frequency $v=\dfrac{c}{\lambda }=\dfrac{3X{{10}^{8}}m/s}{1285X{{10}^{-9}}m}$ --- (2)
Where, the transition observed = 1285nm, c= velocity of light
From equation (1) and (2), the both equations represent the frequency, then both values will resemble equal values.
Equation (1) = equation (2)
\[v=3.29X{{10}^{15}}(Hz)[1/{{3}^{2}}-1/{{n}^{2}}]\]
= $\dfrac{3X{{10}^{8}}}{1285X{{10}^{-9}}}$
Therefore, $\dfrac{1}{{{n}^{2}}}=\dfrac{1}{9}-\dfrac{3X{{10}^{8}}}{1285X{{10}^{-9}}}X\dfrac{1}{3.29X{{10}^{15}}}$ = 0.111-0.071=0.04
Hence, $\dfrac{1}{{{n}^{2}}}=\dfrac{1}{25}\Rightarrow n=5$
So, the value of the transition which is observed at 1285nm is n=5
The radiation corresponding to infrared region is 700nm to millimeter (mm)
The value of the transition n=5 is 1285nm in the infrared region.
Finally the region observed for the given transition is the infrared region.

Note: The longest wavelength in the Paschen series, ${{n}_{1}}=3\And {{n}_{2}}=4$ . Hence, longest wavelength in this series is 18752 ${{A}^{o}}$ . the shortest wavelength in the Paschen series of spectral line in the hydrogen atom is 82.1 ${{A}^{o}}$. The spectral line for the Lyman series is transition state from $n\ge 2\to n=1$ state, similarly the Paschen series are transitions from $n\ge 4\to n=3$ state.