Question

What is the wavelength of the first visible line in the spectrum of doubly ionized lithium? Begin by writing down the formula for the energy levels of the electron in doubly ionized lithium - then consider energy-level differences that give energies in the appropriate (visible) range. Express the answer in terms of "the transition from state \(n\) to state \(n^{\prime}\) is the first visible, with wavelength \(X\)."

Short answer

Answer: The first visible line in the spectrum of doubly ionized lithium has a wavelength of 656.3 nm and corresponds to a transition between the energy levels n = 1 and n' = 3.

Step-by-step solution

Determine the electron configuration of doubly ionized lithium

Doubly ionized lithium means that two electrons have been removed from the lithium atom. The lithium atom has 3 electrons, so the remaining electron will be in the ground state, which is the 1s state. The electron configuration in doubly ionized lithium is just 1 electron in the 1s orbital.

Write down the formula for energy levels of electrons in hydrogen-like ions

The energy levels of electrons in hydrogen-like ions can be determined using the following formula: \(E_n = -\cfrac{Z^2hcR}{n^2}\) Where \(E_n\) represents the energy of the electron corresponding to the principal quantum number \(n\) (in units of joules), \(Z\) represents the atomic number of the ion, \(h\) is the Planck constant, \(c\) is the speed of light, and \(R\) is the Rydberg constant for hydrogen (approximately 1.097 × \(10^{7} \,\text{m}^{-1}\)). For doubly ionized lithium, \(Z = 3\) and \(n\) can take on integer values from 1 to infinity.

Calculate the energy difference between energy levels for doubly ionized lithium

To find the energy difference between energy levels for doubly ionized lithium, we can use the above formula and subtract the energies for the two different principal quantum numbers: \(E_{n' \rightarrow n} = E_{n'} - E_n = \cfrac{Z^2hcR}{n^2} - \cfrac{Z^2hcR}{n'^2}\) We need to find the energy difference that corresponds to a transition in the visible range of the spectrum (400-700 nm).

Determine the energy difference required using the visible range and the Rydberg formula

The energy difference required for a transition in the visible range can be found using the Rydberg formula for the wavelength: \(\cfrac{1}{\lambda} = R \left (\cfrac{1}{n^2} - \cfrac{1}{n'^2} \right )\) With the visible range being between 400 to 700 nm, we must find the pair of principal quantum numbers, \(n\) and \(n'\), that correspond to this range.

Calculate the wavelength of the first visible line in the spectrum of doubly ionized lithium

To calculate the wavelength, we can rearrange the Rydberg formula for the wavelength and substitute the values for \(R\) and \(Z\): \(\lambda = \cfrac{n^2n'^2}{(n'^2 - n^2)RZ^2}\) Now, we can test different values of \(n\) and \(n'\) to find the first visible line. For example, we can start with values of \(n = 1\) and \(n' = 2\). \(\lambda = \cfrac{(1^2)(2^2)}{(2^2 - 1^2)(1.097 × 10^{7} \,\text{m}^{-1})(3^2)} \approx 121.6\,\text{nm}\) This wavelength is outside the visible range, so we can try \(n' = 3\): \(\lambda = \cfrac{(1^2)(3^2)}{(3^2 - 1^2)(1.097 × 10^{7} \,\text{m}^{-1})(3^2)} \approx 656.3\,\text{nm}\) The wavelength of 656.3 nm is within the visible range. So, the first visible line in the spectrum of doubly ionized lithium corresponds to a transition between the energy level \(n = 1\) and \(n' = 3\): "The transition from state \(1\) to state \(3\) is the first visible, with wavelength \(656.3\,\text{nm}\).

Key concepts

Energy Levels of Electrons

Electrons within an atom reside in areas known as energy levels or shells, each with a distinctly quantized energy. An analogy to consider is a ladder, where electrons can only stand on the rungs, not between them. The energy levels follow a predictable pattern, growing further apart the closer they are to the nucleus, and becomes closer as they move away from it. Importantly, each level can only hold a certain number of electrons due to the Pauli exclusion principle. When we consider the energy levels of electrons in hydrogen-like ions, which includes doubly ionized lithium, the energy for each level is given by the formula \[\begin{equation}E_n = -\frac{Z^2hcR}{n^2}\end{equation}\]where the negative sign indicates that the electron is bound to the nucleus; a free electron would have zero potential energy. In these ions, the sole electron can move between energy levels by absorbing or emitting a photon. The energy of the photon corresponds to the energy difference between the initial and final levels. In essence, this energy difference is responsible for the spectral lines we see when observing the electromagnetic spectrum emitted by such ions.

Visible Light Spectrum

The visible light spectrum is a small window of the electromagnetic spectrum that human eyes can detect. It ranges approximately from 400 nm (violet) to 700 nm (red). When an electron transitions between energy levels, it can emit a photon within this range, producing visible light. For an emission to be visible, the energy difference between the higher and lower electron energy levels must correspond to the energy of a photon in the visible spectrum.The colors we see represent different energies; violet photons have the most energy, and red photons have the least. This is why the question about the first visible line in the spectrum of doubly ionized lithium is significant, as it relates to identifying which electron transition corresponds to a wavelength that our eyes can perceive. The electron's transition from a higher to a lower level that emits a photon in the visible spectrum produces a line that can be observed as colored light in experiments or applications like spectroscopy.

Rydberg Formula

The Rydberg formula is quintessential in atomic physics for calculating the wavelengths of spectral lines of many chemical elements. Originally formulated for hydrogen, the expression is\[\begin{equation}\frac{1}{\lambda} = R \left(\frac{1}{n^2} - \frac{1}{n'^2}\right)\end{equation}\]where \begin{align*}\lambda &\text{ is the wavelength},R &\text{ is the Rydberg constant},n &\text{ and } n' \text{ are the principal quantum numbers}.\end{align*}This formula gives us the reciprocal of the wavelength (which is directly proportional to the energy difference between two levels) for an electron transitioning between energy levels. In the exercise, by using known values for the constants and the appropriate quantum numbers, we can determine the wavelength of light emitted by doubly ionized lithium when an electron falls from a higher to a lower energy level. For example, a calculation shows that an electron falling from energy level 3 to 1 would emit light with a wavelength falling into the visible spectrum, specifically in the red color range.

Quantum Numbers

Quantum numbers are the sets of numerical values that provide important details about the quantum state of an electron in an atom, like its distribution around the nucleus, energy, and other properties. There are four quantum numbers:

1. The principal quantum number (n) indicates the main energy level or shell.
2. The azimuthal quantum number (l) describes the shape of the orbital.
3. The magnetic quantum number (m) describes the orientation of the orbital in space.
4. The spin quantum number (s) describes the spin orientation of the electron.

In the context of the textbook exercise, we primarily deal with the principal quantum number, which directly relates to the energy level of an electron in a hydrogen-like ion. This number can take on positive integer values (1, 2, 3, ...), where n=1 denotes the ground state, the lowest energy state an electron can occupy, and n>1 denotes excited states. The greater the value of n, the higher the energy level and the farther from the nucleus the electron's probable location, which affects the possible transitions and resulting spectral lines observed.