The longest wavelength associated with Paschen series is: (Given $R_H = 1.097 \times 10^7$ SI unit)
- $1.094 \times 10^{-6}$ m
- $2.973 \times 10^{-6}$ m
- $3.646 \times 10^{-6}$ m
- $1.876 \times 10^{-6}$ m
The Correct Option is D
Approach Solution - 1
For the longest wavelength in the Paschen series:
\[\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]\]
For the longest wavelength, \( n_1 = 3 \) and \( n_2 = 4 \).
\[\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]\]
\[\frac{1}{\lambda} = R \left[ \frac{1}{9} - \frac{1}{16} \right]\]
\[\frac{1}{\lambda} = R \cdot \frac{16 - 9}{144} = R \cdot \frac{7}{144}\]
Now, substitute \( R = 1.097 \times 10^7 \):
\[\frac{1}{\lambda} = \frac{7 \times 1.097 \times 10^7}{144}\]
\[\lambda = \frac{144}{7 \times 1.097 \times 10^7} = 1.876 \times 10^{-6} \, \text{m}\]
Approach Solution -2
To determine the longest wavelength associated with the Paschen series, we should first understand the concept of spectral lines in the hydrogen atom. The Paschen series corresponds to transitions where the electron falls to the third energy level (\(n_2 = 3\)) from a higher energy level (\(n_1 \gt 3\)).
The formula to calculate the wavelength (\(\lambda\)) of the emitted photon for a transition from level \(n_1\) to \(n_2\) is given by:
\(\frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)\)
Here, \(R_H\) is the Rydberg constant, approximately equal to \(1.097 \times 10^7\) m-1.
To find the longest wavelength in the Paschen series, we need to consider the transition from the smallest possible value of \(n_1\) to \(n_2 = 3\). The smallest possible integer for \(n_1\) is 4 because \(n_1\) must be greater than 3.
Using \(n_1 = 4\) and \(n_2 = 3\), let's calculate the wavelength:
\(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)\)
Calculate the expression inside the parentheses:
\(\frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144}\)
Thus,
\(\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}\)
Calculate this value to find \(\lambda\):
\(\lambda = \frac{144}{1.097 \times 10^7 \times 7} \approx 1.87568 \times 10^{-6} \text{ meters}\)
Rounding this, we get approximately \(1.876 \times 10^{-6} \text{ m}\), which matches the correct answer.
Therefore, the longest wavelength associated with the Paschen series is 1.876 × 10-6 m.
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