Question

The longest wavelength associated with the Paschen series is: \[ \text{(Given } R_H = 1.097 \times 10^7 \, \text{SI unit)}. \]

• A. \( 1.094 \times 10^{-6} \, \text{m} \)
• B. \( 2.973 \times 10^{-6} \, \text{m} \)
• C. \( 3.646 \times 10^{-6} \, \text{m} \)
• D. \( 1.876 \times 10^{-6} \, \text{m} \)

Hint:

To calculate the longest wavelength in a spectral series, use \( n_2 = n_1 + 1 \) and carefully insert the values into the formula for accurate results.

Answer 1

The Correct Option is D

The mathematical expression for the wavelength in the Paschen series is: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \] where: \( n_1 = 3 \), representing the Paschen series, \( n_2 = 4 \), indicating the next energy level above \( n_1 \) for the longest wavelength. By substituting the values: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right). \] Simplification yields: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{9} - \frac{1}{16} \right). \] \[ \frac{1}{\lambda} = R_H \cdot \frac{16 - 9}{144}. \] \[ \frac{1}{\lambda} = R_H \cdot \frac{7}{144}. \] By substituting value of \( R_H = 1.097 \times 10^7 \): \[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot \frac{7}{144}. \] \[ \frac{1}{\lambda} = \frac{7.679 \times 10^7}{144}. \] \[ \lambda = \frac{144}{7.679 \times 10^7} \approx 1.876 \times 10^{-6} \, \text{m}. \] Final Answer: \[ \boxed{1.876 \times 10^{-6} \, \text{m}} \]