Question

The He ion contains only one electron and is therefore a hydrogen-like ion. Calculate the wavelengths, in increasing order, of the first four transitions in the Balmer series of the \(\mathrm{He}^{+}\) ion. Compare these wavelengths with the same transitions in an \(\mathrm{H}\) atom. Comment on the differences. (The Rydberg constant for He is \(4.39 \times 10^{7} \mathrm{~m}^{-1}\).)

Short answer

The He+ ion has longer Balmer series wavelengths than H, due to its higher Rydberg constant.

Step-by-step solution

Understand the Balmer Series

The Balmer series involves transitions where an electron falls to the second energy level (n=2) from a higher energy level (n>2) in a hydrogen-like ion. For He+, we will consider transitions from n=3, 4, 5, and 6.

Use the Rydberg Formula for Wavelength

The wavelength \( \lambda \) for a transition can be calculated using the formula:\\[ \frac{1}{\lambda} = R_Z \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]where \( R_Z \) is the Rydberg constant for the ion, \(n_1 = 2\) (for Balmer series), and \(n_2 = 3, 4, 5, 6\) for the first four transitions. For He+, \( R_Z = 4.39 \times 10^{7} \mathrm{m}^{-1} \).

Calculate Wavelengths for Each Transition

Calculate the wavelengths for each transition:- Transition n=3 to n=2: \[ \frac{1}{\lambda_{32}} = 4.39 \times 10^{7} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ \lambda_{32} \approx 1640 \mathrm{nm} \\]- Transition n=4 to n=2: \[ \frac{1}{\lambda_{42}} = 4.39 \times 10^{7} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] \[ \lambda_{42} \approx 1215 \mathrm{nm} \\]- Transition n=5 to n=2: \[ \frac{1}{\lambda_{52}} = 4.39 \times 10^{7} \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \] \[ \lambda_{52} \approx 1034 \mathrm{nm} \\]- Transition n=6 to n=2: \[ \frac{1}{\lambda_{62}} = 4.39 \times 10^{7} \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \] \[ \lambda_{62} \approx 910 \mathrm{nm} \\]

Compare with Hydrogen Atom

For hydrogen, \( R = 1.097 \times 10^{7} \mathrm{m}^{-1} \). Calculating similarly, the wavelengths for \(H\) for n=3, 4, 5, 6 to n=2 are approximately 656 nm, 486 nm, 434 nm, and 410 nm respectively. The transitions for \(He^+\) are at higher wavelengths than those in \(H\) due to the higher Rydberg constant, which reflects the greater nuclear charge exerting a stronger force on the electron.

Key concepts

Hydrogen-like Ion

A hydrogen-like ion is a fascinating concept in atomic physics. It refers to ions that possess only one electron orbiting around a single nucleus, much like a hydrogen atom. The only difference between them is the nuclear charge. For the hydrogen atom, the nucleus has just one proton, whereas hydrogen-like ions have additional protons. Thus, for helium ions (He\(^{+}\)), the nucleus contains two protons.
This added charge influences the behavior of the electron, such as its energy levels and transitions. Here are some key features of hydrogen-like ions:

• They exhibit spectral characteristics similar to hydrogen due to having a single electron.
• Their energy levels are influenced by the nuclear charge, leading to different transition wavelengths compared to hydrogen.
• They provide excellent systems for exploring basic principles of quantum mechanics and spectroscopy.

These ions, therefore, serve as a simplified model to understand more complex atoms and their electronic transitions.

Rydberg Formula

The Rydberg formula is a fundamental equation in quantum mechanics used to predict the wavelengths of light resulting from electron transitions between energy levels in an atom. Specifically, it helps to determine the spectral lines of hydrogen and hydrogen-like ions. The formula is:\[ \frac{1}{\lambda} = R_Z \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \(\lambda\) is the wavelength, \(R_Z\) is the Rydberg constant specific to the ion in question, and \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and higher energy levels respectively.
This formula is extremely valuable because:

• It allows precise calculations of emission spectra, which are unique to each element and ion.
• The Rydberg constant varies with the nuclear charge, affecting the resulting spectral lines.
• It links the quantum world with observable phenomena, enabling deeper insights into atomic structure.

By applying this equation, scientists can gain insights into the spectral properties of various elements and their ions.

Energy Transition

Energy transitions occur when electrons move between different energy levels in an atom or ion. In hydrogen-like ions, these transitions lead to either the absorption or emission of photons, which are detected as distinct spectral lines. The Balmer series is a prime example of such transitions, involving electrons falling to the second energy level from higher ones.These transitions can be characterized by:

• Specific energy changes: The difference in energy levels determines the energy, and thus the wavelength, of the emitted or absorbed photon.
• Quantum mechanical rules: Only certain transitions are allowed by the quantum mechanics governing electrons.
• Use of the Rydberg formula: This helps calculate the specific wavelengths you’d observe in spectral analysis.

In the He\(^{+}\) ion, analyzing these transitions provides insights into the structure and properties of atoms with single electrons.

He+ Ion Wavelengths

When analyzing the He\(^{+}\) ion, particular focus is on the wavelengths of light emitted during electron transitions, especially in the Balmer series. These wavelengths are crucial for understanding the ion's spectral characteristics, which are determined by its unique energy state configuration due to the nuclear charge. For helium ions, using the Rydberg formula with a Rydberg constant of \(4.39 \times 10^{7} \, \mathrm{m}^{-1}\), we can calculate the first four wavelengths as electrons fall to the \(n=2\) energy level from \(n=3, 4, 5, and n=6\):

• \(n=3 \rightarrow n=2\): approximately \(1640 \, \mathrm{nm}\)
• \(n=4 \rightarrow n=2\): approximately \(1215 \, \mathrm{nm}\)
• \(n=5 \rightarrow n=2\): approximately \(1034 \, \mathrm{nm}\)
• \(n=6 \rightarrow n=2\): approximately \(910 \, \mathrm{nm}\)

These wavelengths are higher than those observed in hydrogen due to the increased nuclear charge, which creates a stronger attractive force, impacting the energy levels the electrons occupy.