Chapter 3: Problem 6
What is the wavelength of the \(2 \alpha\) transition in singly ionized helium? (Hint: The difference between this case and hydrogen is that the charge on the helium nucleus is twice that for hydrogen. Ignore the difference in reduced masses.
Short Answer
Expert verified
The wavelength of the \2 \alpha\ transition in singly ionized helium is approximately 164 nm.
Step by step solution
01
Understand the Rydberg Formula for Hydrogen-like Atoms
The wavelength of spectral lines in hydrogen-like atoms is given by: \[\frac{1}{\text{λ}} = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] where \R\ is the Rydberg constant (1.097 \times 10^7 m^{-1}), \Z\ is the nuclear charge (2 for helium), \_1\ is the lower energy level, and \_2\ is the higher energy level. For the \2 \alpha\ line, \_1 = 2\ and \_2 = 3\.
02
Substitute Values into the Rydberg Formula
Plug the values for ionized helium into the Rydberg formula: \[\frac{1}{\lambda} = 1.097 \times 10^7 m^{-1} \(2^2\) \left( \frac{1}{2^2} - \frac{1}{3^2}\right)\] \[\frac{1}{\lambda} = 1.097 \times 10^7 m^{-1} \times 4 \left( \frac{1}{4} - \frac{1}{9}\right)\]
03
Simplify the Expression Inside the Parentheses
Simplify the term inside the parentheses: \[\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}\] Now plug this back into the formula: \[\frac{1}{\lambda} = 4 \times 1.097 \times 10^7 m^{-1} \times \frac{5}{36}\]
04
Calculate the Value of \(\frac{1}{\lambda}\)
Calculate \[\frac{1}{\lambda}\]: \[\frac{1}{\lambda} = 4 \times 1.097 \times 10^7 m^{-1} \times \frac{5}{36} = 6.0961 \times 10^6 m^{-1}\] Solve for \lambda\: \[\lambda = \frac{1}{6.0961 \times 10^6 m^{-1}} \approx 1.64 \times 10^{-7} m = 164 nm\]
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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 vital tool in atomic physics for calculating the wavelengths of spectral lines in hydrogen-like atoms. It is expressed as follows: ewline ewline ewline ewline ewline \[\frac{1}{\text{λ}} = RZ^2 \bigg(\frac{1}{n_1^2} - \frac{1}{n_2^2}\bigg)\] In this equation:
- \(R\) is the Rydberg constant, approximately equal to \(1.097 \times 10^7 m^{-1}\).
- \(Z\) is the nuclear charge, which is 2 for helium (as helium has twice the charge of hydrogen).
- \(n_1\) and \(n_2\) are the principal quantum numbers of the energy levels from which the electron transitions.
Spectral lines
Spectral lines are unique fingerprints of elements that appear when atoms emit or absorb light. These lines correspond to specific wavelengths and are revealed when the light is dispersed through a prism or diffraction grating.
Spectral lines are evidence of quantum transitions between energy levels within atoms. For example, when an electron in a helium atom transitions from a higher energy level to a lower one, it emits a photon. The wavelength of this photon corresponds to a specific spectral line.
Spectral lines can be used to identify elements and their ionization states in distant stars or gas clouds, making them critical tools in fields such as astrophysics and chemistry. In helium, the spectral lines are further apart compared to hydrogen due to the stronger nuclear charge pulling electrons closer to the nucleus.
Spectral lines are evidence of quantum transitions between energy levels within atoms. For example, when an electron in a helium atom transitions from a higher energy level to a lower one, it emits a photon. The wavelength of this photon corresponds to a specific spectral line.
Spectral lines can be used to identify elements and their ionization states in distant stars or gas clouds, making them critical tools in fields such as astrophysics and chemistry. In helium, the spectral lines are further apart compared to hydrogen due to the stronger nuclear charge pulling electrons closer to the nucleus.
Helium ionization
Ionization in helium involves removing one or more electrons from the atom, resulting in positively charged helium ions. Helium is unique because its nucleus contains two protons, resulting in a greater positive charge than hydrogen. This extra charge means that helium’s electrons experience a stronger electrostatic pull compared to hydrogen electrons.
Singly ionized helium (He II) is a helium ion with one electron removed, leaving it with a +1 charge. To remove a second electron and create He III, it requires even more energy.
Such ionization processes are common in high-energy environments like stars, where the temperature is high enough to provide the necessary energy. Ionized helium produces distinct spectral lines, which help in studying the physical conditions of stellar objects.
Singly ionized helium (He II) is a helium ion with one electron removed, leaving it with a +1 charge. To remove a second electron and create He III, it requires even more energy.
Such ionization processes are common in high-energy environments like stars, where the temperature is high enough to provide the necessary energy. Ionized helium produces distinct spectral lines, which help in studying the physical conditions of stellar objects.
Atomic transitions
Atomic transitions occur when an electron jumps between different energy levels within an atom. These transitions involve the absorption or emission of photons, and the energy (or equivalently, the wavelength) of these photons is determined by the difference in energy between the two levels.
For example, in singly ionized helium, when an electron jumps from the n=3 level to the n=2 level, it emits a photon that corresponds to the 2α line in the helium spectrum. Using the Rydberg formula, this photon's wavelength can be calculated.
Quantum mechanics governs these transitions, ensuring that only specific, quantized energy levels are allowed within atoms. This quantization explains why spectral lines are discrete and provide unique identifiers for each element.
For example, in singly ionized helium, when an electron jumps from the n=3 level to the n=2 level, it emits a photon that corresponds to the 2α line in the helium spectrum. Using the Rydberg formula, this photon's wavelength can be calculated.
Quantum mechanics governs these transitions, ensuring that only specific, quantized energy levels are allowed within atoms. This quantization explains why spectral lines are discrete and provide unique identifiers for each element.