Question

Doubly ionized lithium. \(\mathrm{Li}^{2+}\), abvorbs a photon and juinps from the ground state to its \(n=2\) level. What was the wavelength of the photon?

Short answer

The wavelength of the photon absorbed by the doubly ionized lithium (\(\mathrm{Li}^{2+}\)) atom jumping from the ground state to its \(n=2\) level is \(51.1 nm\).

Step-by-step solution

Identify the atomic number of Lithium

\(\mathrm{Li}^{2+}\) is lithium which has an atomic number Z of 3.

Substitute values into Rydberg formula

Substitute R equals to \(1.097 \times 10^{7} \, m^{-1}\), Z equals to 3, \(n_1\) equals to 1 (ground level) and \(n_2\) equals to 2 into the Rydberg formula: \(\frac{1}{\lambda} = 1.097 \times 10^{7} \times (3)^2 \times (\frac{1}{1^2} - \frac{1}{2^2})\).

Simplify to find \(1/ \lambda\)

Solve to find value for \(\frac{1}{\lambda}\), which equals to \(1.9588 \times 10^{7} \, m^{-1}\).

Solve for \(\lambda\)

To get \(\lambda\), take the reciprocal of this value: \(\lambda = \frac{1}{1.9588 \times 10^{7}} = 5.11 \times 10^{-8} m\) or in terms of nanometers (\(1m = 10^9 nm\)), that's \(51.1 nm\).

Key concepts

Rydberg Formula and its Application

The Rydberg formula is a critical tool in the world of atomic physics. It helps calculate the wavelengths of light emitted or absorbed during electron transitions in atoms. The formula is expressed as: \[\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]where:

• \(\lambda\) is the wavelength of the photon.
• \(R\) is the Rydberg constant, approximately equal to \(1.097 \times 10^7\, \text{m}^{-1}\).
• \(Z\) is the atomic number.
• \(n_1\) and \(n_2\) represent the initial and final energy levels the electron jumps between.

By substituting the given values for a doubly ionized lithium atom \(\text{Li}^{2+}\), with an atomic number \(Z=3\), and transition levels \(n_1=1\) to \(n_2=2\), the formula calculates the inverse of the wavelength, which can then be converted to find the actual wavelength. This calculation gives insight into the energy and characteristics of the photon involved in the transition.

Understanding Atomic Transitions

Atomic transitions describe the movement of electrons between different energy levels or shells within an atom. When an electron absorbs energy, it jumps to a higher energy level. Conversely, when it releases energy, it falls back to a lower level.These absorptions and emissions are not random. They happen at specific energy values and result in the absorption or emission of photons with precise wavelengths. The wavelengths emitted or absorbed are determined by the difference in energy between the two levels involved in the transition. This concept is fundamental to understanding how the Rydberg formula works, as the formula effectively captures the energy change during such transitions to provide the corresponding photon wavelength.For lithium, involving a transition from the ground state \(n=1\) to the first excited state \(n=2\), it requires understanding that a specific amount of energy must be absorbed from a photon for the electron to make this jump. The precise energy is closely linked to the calculated wavelength using the Rydberg formula.

Lithium Ionization and Its Effects

Lithium ionization involves removing electrons from an atom, resulting in a positively charged ion. Specifically, a doubly ionized lithium atom \(\text{Li}^{2+}\) has lost two electrons.Ionization has critical effects on the atomic structure and behavior:

• It increases the effective nuclear charge experienced by the remaining electrons because fewer electrons mean less repulsion.
• This increased effective charge leads to a tighter binding of the remaining electrons, altering the energies of the electronic transitions.
• The energies, and thus wavelengths calculated for these transitions, will be different from those of a neutral atom due to this altered state.

In our example of a doubly ionized lithium atom, the Rydberg formula shows us how ionization significantly impacts the calculations related to electronic transitions. The increased atomic number \(Z=3\) (since only one electron is left) dramatically adjusts the wavelength outcomes, highlighting the effect of ionization on atomic behavior and spectral properties.