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Chapter 27: Problem 44

The Paschen series in the hydrogen emission spectrum is formed by electron transitions from \(n_{\mathrm{i}}>3\) to \(n_{\mathrm{f}}=3.\) (a) What is the longest wavelength in the Paschen series? (b) What is the wavelength of the series limit (the lower bound of the wavelengths in the series)? (c) In what part or parts of the EM spectrum is the Paschen series found (IR, visible, UV, etc.)?

Short Answer

Expert verified
Answer: The key characteristics of the Paschen series in the hydrogen spectrum are (a) the longest wavelength is approximately 1.87 x 10^-6 m (1870 nm), (b) the shortest wavelength (series limit) is approximately 8.22 x 10^-7 m (822 nm), and (c) the series falls entirely within the infrared (IR) region of the electromagnetic spectrum.

Step by step solution

01

Recall the Rydberg formula for hydrogen

Let's recall the Rydberg formula for hydrogen, which is given by: $$\frac{1}{\lambda} = R_H \left(\frac{1}{n_{\mathrm{f}}^2} - \frac{1}{n_{\mathrm{i}}^2}\right)$$ where \(\lambda\) is the wavelength of the emitted light, \(R_H\) is the Rydberg constant for hydrogen, (\(R_H \approx 1.097 \times 10^7 m^{-1}\)), \(n_{\mathrm{f}}\) is the final energy level and \(n_{\mathrm{i}}\) is the initial energy level.
02

Find the longest wavelength in the Paschen series

Since we are asked for the longest wavelength in the Paschen series, we need to consider the smallest possible difference between the energy levels. As the electron transitions to \(n_{\mathrm{f}}=3\), the smallest energy level difference occurs when \(n_{\mathrm{i}}=4\). We will plug these values into the Rydberg formula: $$\frac{1}{\lambda_{\mathrm{longest}}} = R_H \left(\frac{1}{3^2} - \frac{1}{4^2}\right)$$ Now we can solve for \(\lambda_{\mathrm{longest}}\).
03

Calculate the longest wavelength

Solve for \(\lambda_{\mathrm{longest}}\): $$\lambda_{\mathrm{longest}} = \frac{1}{R_H \left(\frac{1}{3^2} - \frac{1}{4^2}\right)}$$ $$\lambda_{\mathrm{longest}} = \frac{1}{1.097 \times 10^7 \left(\frac{1}{9} - \frac{1}{16}\right) m^{-1}}$$ $$\lambda_{\mathrm{longest}} \approx 1.87 \times 10^{-6} m$$
04

Find the shortest wavelength (series limit) in the Paschen series

To find the series limit, consider the largest possible difference between the energy levels, which occurs as \(n_{\mathrm{i}} \rightarrow \infty\). Plug these values into the Rydberg formula: $$\frac{1}{\lambda_{\mathrm{shortest}}} = R_H \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right)$$ Now we can solve for \(\lambda_{\mathrm{shortest}}\).
05

Calculate the shortest wavelength (series limit)

Solve for \(\lambda_{\mathrm{shortest}}\): $$\lambda_{\mathrm{shortest}} = \frac{1}{R_H \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right)}$$ $$\lambda_{\mathrm{shortest}} = \frac{1}{1.097 \times 10^7 \left(\frac{1}{9}\right) m^{-1}}$$ $$\lambda_{\mathrm{shortest}} \approx 8.22 \times 10^{-7} m$$
06

Determine the EM spectrum region of the Paschen series

Based on the longest and shortest wavelengths we calculated, we can determine that the Paschen series falls entirely within the infrared (IR) region of the electromagnetic spectrum (700 nm to 1 mm). The answers are: (a) The longest wavelength in the Paschen series is approximately \(1.87 \times 10^{-6} m\) or 1870 nm. (b) The wavelength of the series limit (the lower bound of the wavelengths in the series) is approximately \(8.22 \times 10^{-7} m\) or 822 nm. (c) The Paschen series is found in the infrared (IR) region of the electromagnetic spectrum.

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