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Chapter 10: Problem 3

The \(\mathrm{Li}^{2+}\) ion is hydrogenic and has a Lyman series at \(740747 \mathrm{~cm}^{-1}, 877\) \(924 \mathrm{~cm}^{-1}, 925933 \mathrm{~cm}^{-1}\), and beyond. Show that the energy levels are of the form \(-h c R / n^{2}\) and find the value of \(R\) for this ion. Go on to predict the wavenumbers of the two longest-wavelength transitions of the Balmer series of the ion and find the ionization energy of the ion.

Short Answer

Expert verified
The Rydberg constant \( R \) specific for the \( Li^{2+} \) ion can be found by simultaneous solutions of the equations for provided Lyman series wavenumbers. Balmer series wavenumbers would subsequently be calculated using the Rydberg formula with \( n_2 = 2 \), and the ionization energy by the energy level formula for a single electron at ground state.

Step by step solution

01

Analyze the Rydberg formula for hydrogenic ions

The Rydberg formula for the spectral lines of a hydrogenic ion is given by \[ \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] where \( \lambda \) is the wavelength, \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively. An alternate form is in terms of wavenumber \( u \) expressed as \[ u = \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] Since we are dealing with wavenumbers, we will use this alternate form for our calculations.
02

Determine the principal quantum numbers

For the Lyman series, the lower energy level \( n_1 \) is always 1, and the higher energy levels \( n_2 \) are 2, 3, 4, and so on. Given the wavenumbers for the Lyman series, we assume they correspond to transitions from the energy levels 2, 3, and 4 to the ground state \( n_1 = 1 \). So, we can write three equations corresponding to each of the given wavenumbers: \[u_1 = RZ^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right)\] \[u_2 = RZ^2\left(\frac{1}{1^2} - \frac{1}{3^2}\right)\] \[u_3 = RZ^2\left(\frac{1}{1^2} - \frac{1}{4^2}\right)\] We can now solve these equations to determine the value of \( R \).
03

Calculate the value of the Rydberg constant for this ion

Substitute the given wavenumbers into the equations from Step 2 to find the value of \( R \). We have \( Z = 3 \) for lithium. Taking the first two wavenumbers as examples and substituting them into the first two equations we get: \[740747 = R \cdot 3^2\left(1 - \frac{1}{4}\right) \] \[877924 = R \cdot 3^2\left(1 - \frac{1}{9}\right) \] Solving these two equations simultaneously yields the value of \( R \).
04

Predict the wavenumbers of the Balmer series

The Balmer series corresponds to transitions from higher energy levels to the second principal quantum level \( n_2 = 2 \). The longest wavelength transitions occur for \( n = 3 \) and \( n = 4 \), leading to the first and second lines of the series. Using the Rydberg formula, we calculate the wavenumbers for these transitions: \[u_{B1} = RZ^2\left(\frac{1}{2^2} - \frac{1}{3^2}\right)\] \[u_{B2} = RZ^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right)\] where \( u_{B1} \) is the first Balmer transition and \( u_{B2} \) is the second Balmer transition.
05

Find the ionization energy of the ion

The ionization energy is the energy required to remove an electron from the ground state to infinity. For the \( Li^{2+} \) ion, it corresponds to removing an electron from the principal quantum level \( n = 1 \) and is given by the equation: \[ E = -h c R Z^2\frac{1}{n^2} \] By substituting the value of \( R \) found in previous steps, \( Z = 3 \), and \( n = 1 \), we can calculate the ionization energy \( E \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spectral Lines
Spectral lines are unique patterns of light emissions or absorptions observed when substances interact with light. This phenomenon can be seen as distinctive lines in a spectroscope and is fundamental to understanding atomic structure. Each element emits or absorbs light at specific frequencies, corresponding to the differences in energy levels of an atom's electrons. When an electron moves between energy levels, it absorbs or emits photons with energy precisely equal to the difference between these levels. This results in the appearance of spectral lines at certain wavelengths or wavenumbers.

In the context of hydrogen-like ions, such as \( Li^{2+} \) in the exercise, spectral lines are the result of electronic transitions between different principal quantum levels. The series of lines created by these transitions are named after the scientists who discovered them, such as the Lyman and Balmer series in hydrogen. The specific positions of the spectral lines can be predicted using the Rydberg formula, making it a crucial tool for understanding the electronic structure of ions.
Principal Quantum Number
The principal quantum number, represented by \(n\), is a critical quantum mechanical concept that determines the energy level of an electron in an atom. This number can be any positive integer, and each value of \(n\) corresponds to an electron shell or energy level. As \(n\) increases, the electron's energy and its average distance from the nucleus also increase.

In the exercise, we look at the transitions of electrons between different energy levels defined by principal quantum numbers. For instance, the Lyman series involves transitions from higher energy levels to the ground state (\(n_1 = 1\)), while the Balmer series involves transitions down to the second energy level (\(n_1 = 2\)). The value of the principal quantum number fundamentally determines the energy of electrons and, consequently, the wavenumber and wavelength of spectral lines observed in the emission spectrum.
Wavenumber
Wavenumber, often denoted by \(u\), is a measure of the number of wave cycles per unit of distance and is inversely proportional to wavelength \(\lambda\). It is typically expressed in units of inverse centimeters (\(cm^{-1}\)) and is used frequently in spectroscopy to represent the position of spectral lines. The higher the wavenumber, the shorter the wavelength and the higher the photon's energy.

Using the Rydberg formula, which relates wavenumbers to energy level transitions in hydrogen-like ions, we can predict specific spectral lines. In our exercise, details on the Lyman series wavenumbers allowed us to calculate the Rydberg constant for the \(Li^{2+}\) ion. Knowing the wavenumber of specific transitions, such as in the Balmer series, helps identify the corresponding spectral lines, a key aspect of understanding and studying atomic and molecular spectra.
Ionization Energy
Ionization energy is defined as the energy required to remove an electron from the ground state of an isolated gaseous atom or ion to form a cation. It is a crucial element in understanding the binding energy and reactivity of an atom or ion. The larger the ionization energy, the harder it is to remove an electron, implying a more stable electron configuration.

In the exercise, we are asked to find the ionization energy of a \(Li^{2+}\) ion. Ionization energy is linked with the principal quantum number, as it represents the energy to move an electron from the ground state (\(n = 1\)) to an infinitely distant point (where \(n = \infty\) and the electron is completely removed). By applying the Rydberg formula and the previously found Rydberg constant, we can calculate the precise value of this ionization energy for the \(Li^{2+}\) ion, enhancing our understanding of its chemical properties and behavior.

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Most popular questions from this chapter

The spectrum of a star is used to measure its radial velocity with respect to the Sun, the component of the star's velocity vector that is parallel to a vector connecting the star's centre to the centre of the Sun. The measurement relies on the Doppler effect, in which radiation is shifted in frequency when the source is moving towards or away from the observer. When a star emitting electromagnetic radiation of frequency \(v\) moves with a speed s relative to an observer, the observer detects radiation of frequency \(v_{\text {receding }}=v f\) or \(v_{\text {approaching }}=v / f\), where \(f=\\{(1-s / c) l(1+s / c)\\}^{1 / 2}\) and \(c\) is the speed of light. It is easy to see that \(v_{\text {receding }}v\) and an approaching star is characterized by a blue shift of its spectrum with respect to the spectrum of an identical, but stationary source. In a typical experiment, \(\mathrm{v}\) is the frequency of a spectral line of an element measured in a stationary Earth- bound laboratory from a calibration source, such as an arc lamp. Measurement of the same spectral line in a star gives \(v_{\max }\) and the speed of recession or approach may be calculated from the value of \(v\) and the equations above. (a) Three Fe I lines of the star HDE 271182 , which belongs to the Large Magellanic Cloud, occur at \(438.882 \mathrm{~nm}, 441.000 \mathrm{~nm}\), and \(442.020 \mathrm{~nm}\). The same lines occur at \(438.392 \mathrm{~nm}, 440.510 \mathrm{~nm}\), and \(441.510 \mathrm{~nm}\) in the spectrum of an Earth-bound iron arc. Determine whether HDE 271182 is receding from or approaching the Earth and estimate the star's radial speed with respect to the Earth. (b) What additional information would you need to calculate the radial velocity of HDE 271182 with respect to the Sun?

The Zeeman effect is the modification of an atomic spectrum by the application of a strong magnetic field. It arises from the interaction between applied magnetic fields and the magnetic moments due to orbital and spin angular momenta (recall the evidence provided for electron spin by the SternGerlach experiment, Section 9.8). To gain some appreciation for the so-called normal Zeeman effect, which is observed in transitions involving singlet states, consider a pelectron, with \(l=1\) and \(m_{l}=0, \pm 1\). In the absence of a magnetic field, these three states are degenerate. When a field of magnitude \(B\) is present, the degeneracy is removed and it is observed that the state with \(m_{l}\) \(=+1\) moves up in energy by \(\mu_{\mathrm{B}} B\), the state with \(m_{l}=0\) is unchanged, and the state with \(m_{l}=-1\) moves down in energy by \(\mu_{\mathrm{B}} B\), where \(\mu_{\mathrm{B}}=e \hbar / 2 m_{e}=9.274 \times\) \(10^{-24} \mathrm{~J} \mathrm{~T}^{-1}\) is the Bohr magneton (see Section 15.1). Therefore, a transition between a \({ }^{1} \mathrm{~S}_{6}\) term and a \({ }^{1} \mathrm{P}_{1}\) term consists of three spectral lines in the presence of a magnetic field where, in the absence of the magnetic field, there is only one. (a) Calculate the splitting in reciprocal centimetres between the three spectral lines of a transition between a \({ }^{1} \mathrm{~S}_{0}\) term and a \({ }^{1} \mathrm{P}_{1}\) term in the presence of a magnetic field of \(2 \mathrm{~T}\) (where I \(\left.\mathrm{T}=1 \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~A}^{-1}\right)\), (b) Compare the value you calculated in (a) with typical optical transition wavenumbers, such as those for the Balmer series of the \(\mathrm{H}\) atom. Is the line splitting caused by the normal Zeeman effect relatively small or relatively large?

The wavefunction of a many-electron closed-shell atom can expressed as a Slater determinant (Section 10.4b). A useful property of determinants is that interchanging any two rows or columns changes their sign and therefore, if any two rows or columns are identical, then the determinant vanishes.
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