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Chapter 38: Problem 21

The Pfund series results from emission/absorption of photons due to transitions of electrons in a hydrogen atom to/from the \(n=5\) energy level from/to higher energy levels. What are the shortest and longest wavelength photons emitted in the Pfund series? Are any of these photons visible?

Short Answer

Expert verified
Answer: The transition with the smallest energy difference that produces the longest wavelength occurs when the electron moves from n=6 to n=5, with a wavelength of approximately 2270 nm. This photon is not visible as it is in the infrared region of the electromagnetic spectrum. The transition with the largest energy difference that produces the shortest wavelength occurs when the electron moves from n=∞ to n=5, with a wavelength of approximately 457 nm. This photon is visible, falling in the blue-violet region of the visible spectrum.

Step by step solution

01

Recall the Rydberg Formula

In order to find the shortest and longest wavelength photons emitted in the Pfund series, we need to use the Rydberg formula for hydrogen, which is given by: $$ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$ Where: - \(\lambda\) is the wavelength of the emitted photon, - \(R_H\) is the Rydberg constant for hydrogen, approximately equal to \(1.097\times10^7 \text{ m}^{-1}\), - \(n_1\) is the initial quantum number (for the Pfund series, \(n_1 = 5\)), - \(n_2\) is the final quantum number (for the Pfund series, \(n_2 > n_1 = 5\)).
02

Determine Initial and Final Quantum Numbers for Longest and Shortest Wavelengths

To find the longest wavelength, we need the smallest difference in the energies of the initial and final levels. Since energy levels get closer with the increase of the quantum number, we have the smallest energy difference when the electron transitions from \(n_2 = 6\) to \(n_1 = 5\). For the shortest wavelength, we need the largest difference in the energies of the initial and final levels. Since the energy levels converge at infinity, we have the largest energy difference when the electron transitions from \(n_2 = \infty\) to \(n_1 = 5\).
03

Calculate the Longest Wavelength

For the longest wavelength, use the Rydberg formula with \(n_1 = 5\) and \(n_2 = 6\): $$ \frac{1}{\lambda_{long}} = R_H \left( \frac{1}{5^2} - \frac{1}{6^2} \right) $$ Solve for \(\lambda_{long}\): $$ \lambda_{long} = \frac{1}{R_H \left( \frac{1}{5^2} - \frac{1}{6^2} \right)} $$ $$ \lambda_{long} = \frac{1}{1.097\times10^7 \text{ m}^{-1} \left( \frac{1}{25} - \frac{1}{36} \right)} $$ $$ \lambda_{long} \approx 2.27\times10^{-6} \text{ m} = 2270 \text{ nm} $$
04

Calculate the Shortest Wavelength

For the shortest wavelength, use the Rydberg formula with \(n_1 = 5\) and \(n_2 = \infty\): $$ \frac{1}{\lambda_{short}} = R_H \left( \frac{1}{5^2} - \frac{1}{\infty^2} \right) $$ Solve for \(\lambda_{short}\): $$ \lambda_{short} = \frac{1}{R_H \left( \frac{1}{5^2} - \frac{1}{\infty^2} \right)} $$ $$ \lambda_{short} = \frac{1}{1.097\times10^7 \text{ m}^{-1} \left( \frac{1}{25} \right)} $$ $$ \lambda_{short} \approx 4.57\times10^{-7} \text{ m} = 457 \text{ nm} $$
05

Check if the Wavelengths are Visible

Now, compare the calculated wavelengths with the visible spectrum range (approx. 380 nm to 750 nm): The longest wavelength in the Pfund series is approximately 2270 nm, which is outside the visible range and falls in the infrared region of the electromagnetic spectrum. The shortest wavelength in the Pfund series is approximately 457 nm, which is inside the visible range and falls in the blue-violet region of the visible spectrum. Therefore, only the shortest wavelength photon in the Pfund series is visible to the human eye.

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