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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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rydberg Formula
The Rydberg Formula is a fundamental equation used in atomic physics to predict the wavelengths of light emitted or absorbed by electrons transitioning between energy levels in an atom. Specifically, for hydrogen, this formula is expressed as: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Here, \(\lambda\) represents the wavelength of the photon emitted or absorbed, while \(R_H\) is the Rydberg constant for hydrogen. This constant has a value of approximately \(1.097 \times 10^7 \text{ m}^{-1}\). The variables \(n_1\) and \(n_2\) are the initial and final quantum numbers, respectively, with \(n_2 > n_1\) in emission cases.
- When an electron falls from a higher energy level \(n_2\) to a lower one \(n_1\), it releases energy in the form of a photon.
- The energy of the photon corresponds to the difference in energy between these two levels.
- The formula gives us a way to calculate this energy by relating it to the wavelength of the light involved.
Hydrogen Atom
The hydrogen atom is the simplest atom and consists of a single proton in its nucleus and one electron orbiting the nucleus. This simplicity makes it an ideal subject for studying atomic transitions and spectral lines. When discussing the hydrogen atom's interactions with light, we often refer to different electron transitions leading to emission or absorption spectra.
Each transition corresponds to an electron dropping from a higher energy level to a lower energy level. Since the energy levels are quantized, or discrete, the electron can only exist at certain energy levels, defined by quantum numbers.
- Electrons in hydrogen can move between different energy levels, releasing or absorbing photons in the process.
- The energy of these photons corresponds directly to the difference between the initial and final energy levels, as highlighted by the Rydberg formula.
- The photon emissions produce specific spectral lines that help identify atomic structure, such as those seen in various series including the Lyman, Balmer, and Pfund series.
Wavelengths
Wavelengths are critical in understanding photons emitted by atoms during electron transitions. A wavelength is the distance between successive peaks of a wave and is typically measured in nanometers (nm) or meters (m). In the context of atomic physics, wavelengths correspond to the energy of the photons emitted or absorbed by the transition of electrons between energy levels.
- Shorter wavelengths indicate higher energy photons, while longer wavelengths correspond to lower energy photons.
- In the Pfund series for hydrogen, the transition from \(n_2 = \infty\) to \(n_1 = 5\) results in the shortest possible wavelength, corresponding to the highest energy photon in the series.
- The longest wavelength occurs at the smallest energy transition, for example, when an electron moves from \(n_2 = 6\) to \(n_1 = 5\).
Quantum Number
Quantum numbers are values that describe specific properties of electrons in atoms, and they help understand electron configurations and atomic structure. The principal quantum number, usually denoted as \(n\), indicates the main energy level or shell occupied by the electron. It is a positive integer and plays a significant role in electron transitions.
- Higher quantum numbers correspond to larger electron orbits and higher energy levels.
- When calculating wavelengths using the Rydberg formula, the initial and final quantum numbers define the energy difference involved in the electron’s transition.
- In the Pfund series for hydrogen, the initial quantum number \(n_1\) is considered to be 5, indicating electrons falling into this energy level from higher levels.
- Understanding these transitions helps identify the wavelengths and consequently the spectral lines in series such as Pfund, Balmer, and others.