# Solutions 18

- Page ID
- 47402

## S18.1

a) The H atom

$$\left [-\dfrac{h^2}{8\pi^2m}\nabla^2 -\dfrac{e^2}{4\pi\varepsilon_0 r}\right ]\psi(r)=E\psi(r) $$

This equation can be solved analytically.

b) The He^{+} ion

$$\left [-\dfrac{h^2}{8\pi^2m_e}\nabla^2 -\dfrac{2e^2}{4\pi\varepsilon_0 r}\right ]\psi(r)=E\psi(r) $$

This equation can be solved analytically.

c) The He atom

$$\left [-\dfrac{h^2}{8\pi^2m_e}\nabla_{r_1}^2 -\dfrac{h^2}{8\pi^2m_e}\nabla_{r_2}^2 -\dfrac{2e^2}{4\pi\varepsilon_0 r_1}-\dfrac{2e^2}{4\pi\varepsilon_0 r_2}+\dfrac{e^2}{4\pi\varepsilon_0 r_{12}}\right ]\psi(\overrightarrow {r_1},\overrightarrow{r_2})=E\psi(\overrightarrow{r_1},\overrightarrow{r_2}) $$

This equation cannot be solved analytically.

d) The H_{2}^{+} molecule

$$\left [-\dfrac{h^2}{8\pi^2M_1}\nabla_{R_1}^2 -\dfrac{h^2}{8\pi^2M_2}\nabla_{R_2}^2 -\dfrac{h^2}{8\pi^2m_e}\nabla_r^2 -\dfrac{e^2}{4\pi\varepsilon_0 | \overrightarrow{R_1}-\overrightarrow{r} |} -\dfrac{e^2}{4\pi\varepsilon_0 | \overrightarrow{R_2}-\overrightarrow{r}|}+\dfrac{e^2}{4\pi\varepsilon_0 |\overrightarrow{R_1}-\overrightarrow{R_2}|}\right ]\psi(\overrightarrow {R_1},\overrightarrow{R_2}, \overrightarrow{r})=E\psi(\overrightarrow{R_1},\overrightarrow{R_2}, \overrightarrow{r}) $$

M_{1} and M_{2} are the mass of the two protons, \(\overrightarrow{R_1}\) and \( \overrightarrow{R_2}\) are the location vectors of the two protons, m_{e} is the electron mass, \(\overrightarrow{r}\) is the location vector of the electron.

This equation can be solved analytically (although we often just invoke he LCAO approximation anyway).

e) The H_{2} molecule

$$[-\dfrac{h^2}{8\pi^2M_1}\nabla_{R_1}^2 -\dfrac{h^2}{8\pi^2M_2}\nabla_{R_2}^2 -\dfrac{h^2}{8\pi^2m_e}\nabla_{r_1}^2 -\dfrac{h^2}{8\pi^2m_e}\nabla_{r_2}^2 -\dfrac{e^2}{4\pi\varepsilon_0 | \overrightarrow{R_1}-\overrightarrow{r_1} |} -\dfrac{e^2}{4\pi\varepsilon_0 | \overrightarrow{R_2}-\overrightarrow{r_1}|}-\dfrac{e^2}{4\pi\varepsilon_0 | \overrightarrow{R_1}-\overrightarrow{r_2} |} $$

$$ \left[ -\dfrac{e^2}{4\pi\varepsilon_0 | \overrightarrow{R_2}-\overrightarrow{r_2}|} +\dfrac{e^2}{4\pi\varepsilon_0 |\overrightarrow{r_1}-\overrightarrow{r_2}|} +\dfrac{e^2}{4\pi\varepsilon_0 |\overrightarrow{R_1}-\overrightarrow{R_2}|}\right ]\psi(\overrightarrow {R_1},\overrightarrow{R_2}, \overrightarrow{r_1},\overrightarrow{r_2})=E\psi(\overrightarrow{R_1},\overrightarrow{R_2}, \overrightarrow{r_1},\overrightarrow{r_2}) $$

M_{1} and M_{2} are the mass of the two protons, \(\overrightarrow{R_1}\) and \( \overrightarrow{R_2}\) are the location vectors of the two protons, m_{e} is the electron mass, \(\overrightarrow{r_1}\) and \(\overrightarrow{r_2}\) are the location vectors of the two electron.

This equation cannot be solved analytically.

## S18.2

An ionic bond wave function takes into account the probability that the electrons of a multiatomic molecule might exist on the same atom; the covalent bond wave function assumes they exist on separate atoms.

## S18.3

N_{2.}

Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond. N_{2} has a triple bond, O_{2} has a double bond, and F_{2 }has a single bond, therefore, the bond order for N_{2} is the highest and the bond of N_{2} is the strongest, thus it is the shortest bond among the three.

## S18.4

$$\mu=Q \times r=(2\times 1.6\times10^{-19} C) \times(100 pm \times \dfrac{1\times10^{-12} m}{1 pm})=3.2\times10^{-29} C\cdot {m}=3.2\times10^{-29} C\cdot {m}\dfrac{1 D}{3.336\times10^{-30}C\cdot{m}}=9.6 D$$

## S18.5

Substitute \(\phi=N\psi\) into the left hand side of the Schrodinger equation, we get:

$$(\hat{T}+\hat{V})\phi=(\hat{T}+\hat{V})N\psi=N\cdot{ (\hat{T}+\hat{V})}\psi=N\cdot {E\psi}$$

Substitute it to the right hand side of the Schrodinger equation, we get:

$$E\phi=E\cdot{N\psi}=N\cdot{E\psi}$$

Therefore,

$$(\hat{T}+\hat{V})\phi=E\phi$$

This demonstrates that \(\phi\) is also a valid wavefunction to the Schrodinger equation, and many wavefunctions can be generated that are valid solutions to the Schrodinger equation. The way to determine a specific value for N is through normalization, because the total probability of finding a particle is 1, so the integral of \(|\phi|^2\) over the whole space must be 1.

## S18.6

Substitute \(\phi=N_1\psi_1+N_2\psi_2\) into the left hand side of the Schrodinger equation, we get:

$$(\hat{T}+\hat{V})\phi=(\hat{T}+\hat{V})(N_1 \psi_1+N_2 \psi_2)=N_1 \cdot{ (\hat{T}+\hat{V})}\psi_1+N_2 \cdot{ (\hat{T}+\hat{V})}\psi_2=N_1\cdot {E\psi_1}+N_2\cdot {E\psi_2}=E\cdot({N_1\psi_1+N_2\psi_2})=E\cdot\phi$$

Therefore,

$$(\hat{T}+\hat{V})\phi=E\phi$$

This demonstrates that \(\phi\) is a valid wavefunction to the Schrodinger equation. We can determine N_{1} and N_{2} through normalization and boundary conditions.

\(\phi\) can be expanded to three terms of \(\psi\) wavefunctions. It can also be expanded to an infinite number of \(\psi\) wavefunctions and still be a valid wavefunction.