Question

The longest wavelength associated with Paschen series is: (Given $R_H = 1.097 \times 10^7$ SI unit)

• A. $1.094 \times 10^{-6}$ m
• B. $2.973 \times 10^{-6}$ m
• C. $3.646 \times 10^{-6}$ m
• D. $1.876 \times 10^{-6}$ m

Answer 1

The Correct Option is D

To determine the longest wavelength in the Paschen series, the formula is applied:
\[\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]\]
For the longest wavelength, \( n_1 \) is set to 3 and \( n_2 \) to 4.
Substituting these values yields:
\[\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]\]
\[\frac{1}{\lambda} = R \left[ \frac{1}{9} - \frac{1}{16} \right]\]
Simplifying the expression:
\[\frac{1}{\lambda} = R \cdot \frac{16 - 9}{144} = R \cdot \frac{7}{144}\]
Using the value \( R = 1.097 \times 10^7 \):
\[\frac{1}{\lambda} = \frac{7 \times 1.097 \times 10^7}{144}\]
Solving for \( \lambda \):
\[\lambda = \frac{144}{7 \times 1.097 \times 10^7} = 1.876 \times 10^{-6} \, \text{m}\]