Question

The series of emission lines of the hydrogen atom for which \(n_{f}=3\) is called the Paschen series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Paschen series are observed. (b) Calculate the wavelengths of the first three lines in the Paschen series - those for which \(n_{i}=4,5\), and 6 .

Short answer

The Paschen series of the hydrogen atom lies in the infrared region of the electromagnetic spectrum with a minimum wavelength of approximately 820.5 nm. The first three lines in the Paschen series have wavelengths of approximately 1875.6 nm, 2631.1 nm, and 3444.3 nm.

Step-by-step solution

Understand the Rydberg formula

The Rydberg formula is used to calculate the wavelengths of the lines in an atomic spectrum. It is given as: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \) Here, λ is the wavelength, \(R_H\) is the Rydberg constant for hydrogen (\(R_H = 1.097373 \times 10^7\:m^{-1}\)), \(n_f\) is the final principal quantum number, and \(n_i\) is the initial principal quantum number.

Determine the Region of the Electromagnetic Spectrum for the Paschen Series

We know that for the Paschen series, the final principal quantum number (\(n_f\)) is 3. To find the region of the electromagnetic spectrum, let's find the limit of the wavelength when \(n_i\) approaches infinity: \( \lim_{n_i \to \infty} \lambda = \frac{1}{R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) } = \frac{1}{R_H \cdot \frac{1}{n_f^2}} \) Now, substitute the value of the Rydberg constant and \(n_f\): \( \lambda_{min} = \frac{1}{1.097373 \times 10^7\:m^{-1} \cdot \frac{1}{3^2}} \approx 820.5\: nm \) Since the wavelength is around 820.5 nm, the Paschen series lies in the infrared region of the electromagnetic spectrum.

Calculate the Wavelengths of the First Three Lines

To find the wavelengths of the first three lines in the Paschen series, we will use the Rydberg formula and the given values of \(n_i = 4, 5,\) and 6. 1. For \(n_i = 4\): \( \frac{1}{\lambda_1} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \) Calculate the wavelength: \( \lambda_1 = \frac{1}{1.097373 \times 10^7\:m^{-1} \cdot (0.0139)} \approx 1875.6\: nm \) 2. For \(n_i = 5\): \( \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \) Calculate the wavelength: \( \lambda_2 = \frac{1}{1.097373 \times 10^7\:m^{-1} \cdot (0.009444)} \approx 2631.1\: nm \) 3. For \(n_i = 6\): \( \frac{1}{\lambda_3} = R_H \left( \frac{1}{3^2} - \frac{1}{6^2} \right) \) Calculate the wavelength: \( \lambda_3 = \frac{1}{1.097373 \times 10^7\:m^{-1} \cdot (0.006944)} \approx 3444.3\: nm \) So, the first three lines of the Paschen series have wavelengths of approximately 1875.6 nm, 2631.1 nm, and 3444.3 nm.