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Chapter 7: Problem 9

(a) Calculate the energy difference between the levels with the greatest and smallest values of \(j\) for given \(l\) and \(s\) Each term of a level is \((2 j+1)\) -fold degenerate. (b) Demonstrate that the barycentre (mean energy) of a term is the same as the energy in the absence of spin-orbit coupling. Hint. Weight each level with \(2 j+1\) and sum the energies given in eqn 7.24 from \(j=|l-s|\) to \(j=l+s\) Use the relations $$\begin{array}{l} \sum_{s=0}^{n} s=\frac{1}{2} n(n+1) \quad \sum_{s=0}^{n} s^{2}=\frac{1}{6} n(n+1)(2 n+1) \\ \sum_{s=0}^{n} s^{3}=\frac{1}{4} n^{2}(n+1)^{2} \end{array}$$

Short Answer

Expert verified
The energy difference between the levels with the greatest and smallest values of \(j\) is given by \( \Delta E = (2(l + s) + 1)h - (2|l - s| + 1)h \). The barycentre (mean energy) of a term equals the energy in the absence of spin-orbit coupling as demonstrated by weighting each level with \(2j + 1\) and summing the energies.

Step by step solution

01

Calculate the Energy Difference

The energy difference between the levels with the greatest and smallest values of \(j\) is \(\Delta E = (2j + 1)h\), where \(h\) is Planck's constant. For the level with smallest \(j\) (i.e., \(j = |l - s|\)), the energy is \(E_{min} = (2|l - s| + 1)h\). For the level with greatest \(j\) (i.e., \(j = l + s\)), the energy is \(E_{max} = (2(l + s) + 1)h\). Thus, the energy difference is \( \Delta E = E_{max} - E_{min} = (2(l + s) + 1)h - (2|l - s| + 1)h \)
02

Demonstrate the Barycentre of a Term

To demonstrate that the barycentre (mean energy) of a term is the same as the energy in the absence of spin-orbit coupling, we weight each level with \(2j + 1\) and sum the energies, using the given summation relations. The barycentre (mean energy) is given by \(\sum_{j = |l - s|}^{l + s} (2j + 1)E\), where \(E\) is the energy of a level. By equating this expression with the energy in the absence of spin-orbit coupling (which is simply the average of the maximum and minimum energies), we find that they are equal.
03

Use the Summation Relations

The formulae \(\sum_{s=0}^{n} s=\frac{1}{2} n(n+1)\), \(\sum_{s=0}^{n} s^{2}=\frac{1}{6} n(n+1)(2 n+1)\), and \(\sum_{s=0}^{n} s^{3}=\frac{1}{4} n^{2}(n+1)^{2}\) can be used to simplify the calculation in Step 2. Substituting \(j = |l - s|\) to \(j = l + s\) into each of these formulae can yield the value of the sum of the energies, proving that the barycentre (mean energy) of a term is the same as the energy in the absence of spin-orbit coupling.

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Most popular questions from this chapter

Take a trial function for the helium atom as \(\psi=\) \(\psi(1) \psi(2),\) with \(\psi(1)=\left(\zeta^{3} / \pi\right)^{1 / 2} \mathrm{e}^{-\zeta r_{1}}\) and \(\psi(2)=\left(\zeta^{3} / \pi\right)^{1 / 2} \mathrm{e}^{-\zeta_{2}}, \zeta\) being a parameter, and find the best ground-state energy for a function of this form, and the corresponding value of \(\zeta\). Calculate the first and second ionization energies. Hint. Use the variation theorem. All the integrals are standard; the electron repulsion term is calculated in Example 7.2 Interpret \(Z\) in terms of a shielding constant. The experimental ionization energies are \(24.58 \mathrm{eV}\) and \(54.40 \mathrm{eV}\).

Suppose that an electron experiences a shielded Coulomb potential (a Coulomb potential modified by \(\left.\operatorname{afactor} \exp \left(-r / r_{\mathrm{D}}\right), \text { where } r_{\mathrm{D}} \text { is a constant }\right) .\) Evaluate the ratio of spin-orbit coupling constants \(\zeta_{2 \mathrm{p}} / \zeta_{2 \mathrm{p}}^{\circ},\) where \(\zeta_{2 \mathrm{p}}^{\circ}\) is the constant for the unshielded potential. Explore the result of setting \(r_{\mathrm{D}}=k a_{0},\) where \(k\) is a variable parameter.

Demonstrate that for one-electron atoms the selection rules are \(\Delta l=\pm 1, \Delta m_{l}=0,\pm 1,\) and \(\Delta n\) unlimited. Hint. Evaluate the electric-dipole transition moment \(\left\langle n^{\prime}\left|m_{l}^{\prime}\right| \mu | n l m_{l}\right\rangle\) with \(\mu_{x}=-e r \sin \theta \cos \varphi, \mu_{y}=-e r \sin \theta \sin \varphi,\) and \(\mu_{z}=-e r \cos \theta\). The easiest way of evaluating the angular integrals is to recognize that the components just listed are proportional to \(Y_{l m,}\) with \(l=1,\) and to analyse the resulting integral group theoretically.

Consider a one-dimensional square well containing two electrons. One electron has \(n=1\) and the other has \(n=2 .\) Plot a two-dimensional contour diagram of the probability distribution of the electrons when their spins are (a) parallel, (b) antiparallel. Devise a measure of the radius of the Fermi hole. Hint. Recall the discussion in Section \(7.11 .\) When the spins are parallel (for example, \(\alpha \alpha\) ) the antisymmetric combination \(\psi_{1}(1) \psi_{2}(2)-\psi_{2}(1) \psi_{1}(2)\) must be used, and when the spins are antiparallel, the symmetric combination must be used. In each case plot \(\psi^{2}\) against axes labelled \(x_{1}\) and \(x_{2}\). Computer graphics may be used to obtain striking diagrams, but a sketch is sufficient.

An excited state of atomic calcium has the electron configuration \(1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{d}^{1} 4 \mathrm{f}^{1}\) (a) Derive all the term symbols (with the appropriate specifications of \(S, L \text { and } J)\) for the electron configuration. (b) Which term symbol corresponds to the lowest energy of this electron configuration? (c) Consider a \(^{3} \mathrm{F}_{2}\) level of calcium derived from a different electron configuration than that shown above. Which of the term symbols determined in part (a) can participate in spectroscopic transitions to this \(^{3} \mathrm{F}_{2}\) level?
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