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In this problem, you will supply the missing steps in the derivation of the formula \(E_{\text {singlet}}=E_{1 s}+E_{2 s}+J+K\) for the singlet level of the \(1 s^{1} 2 s^{1}\) configuration of He. a. Expand Equation (22.17) to obtain $$\begin{aligned} E_{\text {singlet}}=\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)]\left(\hat{H}_{1}\right) \\ +\frac{1}{2} \iint_{[1 s(1) 2 s(2)+2 s(1) 1 s(2)] d \tau_{1} d \tau_{2}}(10(1) 2 s(2)+2 s(1) 1 s(2)]\left(\hat{H}_{2}\right) \\ +\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)] \times \\ &\left(\frac{e^{2}}{4 \pi \varepsilon_{0}\left|r_{1}-r_{2}\right|}\right) \times \\ &[1 s(1) 2 s(2)+2 s(1) 1 s(2)] d \tau_{1} d \tau_{2} \end{aligned}$$ b. Starting from the equations \(\hat{H}_{i} 1 s(i)=E_{1 s} 1 s(i)\) and \(\hat{H}_{i} 2 s(i)=E_{2 s} 2 s(i),\) show that \(E_{\text {singlet}}=E_{1 s}+E_{2 s}\) $$\begin{array}{c} +\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)]\left(\frac{e^{2}}{4 \pi \varepsilon_{0}\left|r_{1}-r_{2}\right|}\right) \times \\ {[1 s(1) 2 s(2)+2 s(1) 1 s(2)] d \tau_{1} d \tau_{2}} \end{array}$$ c. Expand the previous equation using the definitions $$\begin{array}{c} J=\frac{e^{2}}{8 \pi \varepsilon_{0}} \iint[1 s(1)]^{2}\left(\frac{1}{\left|r_{1}-r_{2}\right|}\right)[2 s(2)]^{2} d \tau_{1} d \tau_{2} \text { and } \\ K=\frac{e^{2}}{8 \pi \varepsilon_{0}} \iint[1 s(1) 2 s(2)]\left(\frac{1}{\left|r_{1}-r_{2}\right|}\right)[1 s(2) 2 s(1)] \times \end{array}$$ to obtain the desired result, \(E_{\text {singlet}}=E_{1 s}+E_{2 s}+J+K\).

Step by step solution

01

Expand Equation (22.17) as given

First we need to expand Equation (22.17) as follows: \(\begin{aligned} E_{\text {singlet}}=\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)]\left(\hat{H}_{1}\right) \\ +\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)]\left(\hat{H}_{2}\right)d \tau_{1} d \tau_{2} \\ +\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)] \times \\ &\left(\frac{e^{2}}{4 \pi \varepsilon_{0}\left|r_{1}-r_{2}\right|}\right) \times \\ &[1 s(1) 2 s(2)+2 s(1) 1 s(2)] d \tau_{1} d \tau_{2} \end{aligned}\)

02

Show that \(E_{\text {singlet}}=E_{1s}+E_{2s}\)

Next, we will use the given equations for \(\hat{H}_{i} 1 s(i)\) and \(\hat{H}_{i} 2 s(i)\), which are: \(\hat{H}_{i} 1 s(i)=E_{1 s} 1 s(i)\) and \(\hat{H}_{i} 2 s(i)=E_{2 s} 2 s(i)\) We will substitute these equations into the expanded equation we obtained in Step 1, and find that: \(E_\text{singlet} = E_{1s} + E_{2s} \\ +\frac{1}{2} \iint[1 s(1) 2 s(2)+2 s(1) 1 s(2)]\left(\frac{e^{2}}{4 \pi \varepsilon_{0}\left|r_{1}-r_{2}\right|}\right) \times \\ {[1 s(1) 2 s(2)+2 s(1) 1 s(2)] d \tau_{1} d \tau_{2}}\)

03

Expand the previous result using the definitions of J and K

Now we need to expand the result from step 2 using the definitions of J and K: \(J=\frac{e^{2}}{8 \pi \varepsilon_{0}} \iint[1 s(1)]^{2}\left(\frac{1}{\left|r_{1}-r_{2}\right|}\right)[2 s(2)]^{2} d \tau_{1} d \tau_{2}\) \(K=\frac{e^{2}}{8 \pi \varepsilon_{0}} \iint[1 s(1) 2 s(2)]\left(\frac{1}{\left|r_{1}-r_{2}\right|}\right)[1 s(2) 2 s(1)] d \tau_{1} d \tau_{2}\) By substituting these expressions into the result from step 2, we finally obtain the desired result: \(E_{\text {singlet}}=E_{1 s}+E_{2 s}+J+K\)

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Key Concepts

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

Singlet State

In quantum chemistry, a singlet state refers to a specific type of electronic state of an atom or molecule where all electron spins are paired. This results in the total spin of the system being zero. The idea of the singlet state is essential because it often relates to the lowest energy state of electronic configurations. This state is particularly important when examining the properties and reactivity of molecules and atoms. In the context of the helium atom, the singlet state corresponds to configurations like the

• 1s², where both electrons occupy the lowest energy orbital,
• 1s¹ 2s¹, where one electron is in the 1s orbital and the other in the 2s orbital.

To comprehend the energy of such a state, it’s crucial to calculate terms like electron repulsion and exchange interactions, as these will influence the overall stability and energy of the atom in the singlet configuration.

Helium Atom

The helium atom is one of the simplest multi-electron systems, consisting of two protons and two electrons. This simplicity makes it a perfect subject of study for quantum mechanical models and calculations, serving as a benchmark for more complex systems. Understanding helium's electron configuration helps in dealing with topics such as electron correlation. In its ground state, helium's electron configuration is 1s², with both electrons occupying the lowest energy level in the spherical 1s orbital. When excited to the

• 1s¹ 2s¹ configuration, one electron moves to a higher energy level,
• creating a situation where electron-electron interactions are even more significant due to their different spatial distribution.

Addressing these interactions through Hartree-Fock theory allows chemists to approximate the wave functions of the helium atom under these circumstances.

Two-electron Integrals

Two-electron integrals are mathematical expressions that arise in the Hartree-Fock method. They represent interactions between two electrons in a given system. These integrals allow us to calculate the potential due to repulsion between electron pairs and how the electrons affect each other’s behavior. In a system like the helium atom, where two electrons move in potential fields created by the nuclei, calculating these integrals is crucial to approximating the electron distribution. The integrals can be complex because they involve multiple coordinates, taking into account

• the spatial relationships between electrons,
• their relative energies,
• and how these factors influence the total energy of the system.

Two-electron integrals are significant for understanding more about electron correlation and spin coupling, integral parts of understanding atomic and molecular systems.

Electron Repulsion Term

When discussing atomic systems, particularly multi-electron ones like helium, the electron repulsion term is a significant component of the total electronic energy. This term accounts for the repulsive force between electrons, which is necessary to consider due to the negatively charged nature of electrons, which inherently repel each other. This term is mathematically expressed as the

• Coulomb repulsion term, which involves integrating over the spatial distribution of the electrons.

In the bid to find the energy of the singlet state in helium, this electron repulsion term becomes part of the expression involving integrals. It’s critical to correctly evaluate the electron repulsion to form an accurate picture of the electronic structure, as this influences everything from

• chemical bonding,
• molecular geometry,
• to atom-electron interaction dynamics.

The Hartree-Fock method helps by providing a way to solve these complex integrals through approximation, making it feasible to determine energies and properties of many-electron systems.