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Chapter 7: Problem 20

\(\mathrm{On}\) the basis of the Thomas-Fermi model of an atom, evaluate the radius within which there is a 50 per cent probability of finding the electron density and evaluate it for the Period 2 elements. Hint. Use the radial distribution function \(r^{2} \rho\).

Short Answer

Expert verified
The solution to this exercise involves understanding the Thomas-Fermi model, radial distribution function and calculation of probability. The exact values for the radius where there is a 50% probability of finding the electron density would depend on the specific form of the radial distribution function \(\rho(r)\) for each Period 2 element.

Step by step solution

01

Understand the Thomas-Fermi Model

The Thomas-Fermi model is a model of an atom in quantum mechanics, which describes the electron cloud surrounding the atomic nucleus. This model investigates how the electrons in an atom are distributed in space.
02

Identify the Radial Distribution Function

In this model, the electron density is given by the radial distribution function, \(r^{2} \rho\). The function \(r^{2} \rho(r)\) gives the probability that the electron will be found in a shell of thickness dr at a distance r from the nucleus.
03

Evaluate the radius of 50% probability

The task is to evaluate the radius within which there is a 50% chance of finding the electron density. This means calculating the radius r such that the integral of \(r^{2} \rho(r)\) from 0 to r equals half of the total probability (which is 1). This results in the inequality \(\int_{0}^{r} r^{2} \rho(r) dr = 0.5\). Solve this inequality using the known form of \(\rho(r)\) and numerical methods.
04

Evaluate for Period 2 elements

Repeat the process described in the previous step for Period 2 elements, i.e. elements with atomic numbers 3 (Lithium) through to 10 (Neon). Remember that the distribution function \(\rho(r)\) will differ for each element, reflecting the different electron configurations.

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Most popular questions from this chapter

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?

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 \(54.40 \mathrm{eV}\).

(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 $$\begin{array}{l} \sum_{s=0}^{n} s=\frac{1}{2} n(n+1) \quad \sum_{s=0}^{n} s^{2}=\frac{1}{6} n(n+1)(2 n+1) \\ \sum_{s=0}^{n} s^{3}=\frac{1}{4} n^{2}(n+1)^{2} \end{array}$$
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