\[ \Psi (r) = \text{exp} \left[ - \alpha \left( r_1 + r_2 \right) \right] \left( 1 + \beta r_{12} \right] \nonumber \] When the wavefunction shown above is used in a variational method calculation for the ground state energy for two-electron atoms or ions the two-parameter equation shown below for the energy is obtained. This equation is then minimized simultaneously with respect to the adjustable parameters, α and b. Nuclear charge: Z = 1 Seed values for scale factors: \( \begin{matrix} \alpha = Z & \beta = .7 \end{matrix} \) Contributions to total energy: \[ \begin{matrix} T ( \alpha,~ \beta ) = \frac{ \frac{1}{2} + \frac{25 \beta}{16 \alpha} + \frac{2 \beta^2}{ \alpha^2}}{ \frac{1}{2 \alpha^2} + \frac{35 \beta}{16 \alpha^3} + \frac{3 \beta^2}{ \alpha^4}} & V_{ne} ( \alpha,~ \beta ) = \frac{ - \frac{Z}{ \alpha} - \frac{15 Z \beta}{4 \alpha^2} - \frac{9Z \beta^2}{ 2 \alpha^3}}{ \frac{1}{2 \alpha^2} + \frac{35 \beta}{16 \alpha^3} + \frac{3 \beta^2}{ \alpha^4}} & V_{ee} ( \alpha,~ \beta ) = \frac{ - \frac{5}{16 \alpha} - \frac{ \beta}{ \alpha^2} - \frac{35 \beta^2}{32 \alpha^3}}{ \frac{1}{2 \alpha^2} + \frac{35 \beta}{16 \alpha^3} + \frac{3 \beta^2}{ \alpha^4}} \end{matrix} \nonumber \] Minimization of the total energy with respect to the variational parameters: \[ \begin{matrix} E ( \alpha,~ \beta ) = T ( \alpha,~ \beta ) + V_{ne} ( \alpha,~ \beta ) + V_{ee} ( \alpha,~ \beta ) & \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \text{Minimize} (E,~ \alpha,~ \beta ) & \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0.8257 \\ 0.4934 \end{pmatrix} & E ( \alpha,~ \beta ) = -0.5088 \end{matrix} \nonumber \] Experimental ground state energy: \[ E_{exp} = -2.9037 \nonumber \] Calculate error in calculation: \[ \begin{matrix} \text{Error} = \left| \frac{E_{exp} - E ( \alpha,~ \beta )}{E_{exp}} \right| & \text{Error} = 82.4782 \% \end{matrix} \nonumber \] Fill in the table and answer the questions below: \[ \begin{pmatrix} \Psi & \text{H} & \text{He} & \text{Li} & \text{Be} \\ \alpha & 0.8257 & 1.8497 & 2.8564 & 3.8592 \\ \beta & 0.4934 & 0.3658 & 0.3354 & 0.3213 \\ E_{atom} & -0.5088 & -2.8911 & -7.2682 & -13.6441 \\ E_{atom} \text{(exp)} & -0.5277 & -2.9037 & -7.2838 & -13.6640 \\ \% \text{Error} & 3.59 & 0.433 & 0.215 & 0.146 \end{pmatrix} \nonumber \] Fill in the table below and explain why this trial wave function gives better results than the previous trial wave function. \[ \begin{pmatrix} \text{WF4} & \text{E} & \text{T} & \text{V}_{ne} & \text{V}_{ee} \\ \text{H} & -0.5088 & 0.5088 & -1.3907 & 0.3731 \\ \text{He} & -2.8911 & 2.8911 & -6.7565 & 0.9743 \\ \text{Li} & -7.2682 & 7.2682 & -16.1288 & 1.5924 \\ \text{Be} & -13.6441 & 13.6441 & -29.5025 & 2.2144 \end{pmatrix} \nonumber \] \[ \begin{matrix} T ( \alpha,~ \beta ) = 0.5088 & V_{ne} ( \alpha,~ \beta ) = -1.3907 & V_{ee} ( \alpha,~ \beta ) = 0.3731 \end{matrix} \nonumber \] Explain the importance of the parameter β. Why does its magnitude decrease as the nuclear charge increases? The parameter β adds weight to the r 12 term which most directly represents electron correlation in the wavefunction. As the nuclear charge increases, as we have previously seen, V ee becomes less important as a percentage of the total energy. Thus, the impact of the electron correlation term becomes less significant. Demonstrate that the virial theorem is satisfied. \[ \begin{matrix} E ( \alpha,~ \beta ) = -0.5088 & -T ( \alpha,~ \beta ) = -0.5088 & \frac{V_{ne} ( \alpha,~ \beta ) + V_{ee} ( \alpha,~ \beta )}{2} = -0.5088 \end{matrix} \nonumber \] Add the results for this wave function to your summary table for all wave functions. \[ \begin{matrix} \begin{pmatrix} \text{H} & \text{E} & \text{T} & \text{V}_{ne} & \text{V}_{ee} \\ \text{WF1} & -0.4727 & 0.4727 & -1.375 & 0.4297 \\ \text{WF2} & -0.4870 & 0.4870 & -1.3705 & 0.3965 \\ \text{WF3} & -0.5133 & 0.5133 & -1.3225 & 0.2958 \\ \text{WF4} & -0.5088 & 0.5088 & -1.3907 & 0.3731 \end{pmatrix} & \begin{pmatrix} \text{He} & \text{E} & \text{T} & \text{V}_{ne} & \text{V}_{ee} \\ \text{WF1} & -2.8477 & 2.8477 & -6.7500 & 1.0547 \\ \text{WF2} & -2.8603 & 2.8603 & -6.7488 & 1.0281 \\ \text{WF3} & -2.8757 & 2.8757 & -6.7434 & 0.9921 \\ \text{WF4} & -2.8911 & 2.8911 & -6.7565 & 0.9743 \end{pmatrix} \\ \begin{pmatrix} \text{Li} & \text{E} & \text{T} & \text{V}_{ne} & \text{V}_{ee} \\ \text{WF1} & -7.2227 & 7.2227 & -16.1250 & 1.6797 \\ \text{WF2} & -7.2350 & 7.2350 & -16.1243 & 1.6544 \\ \text{WF3} & -7.2487 & 7.2487 & -16.1217 & 1.6242 \\ \text{WF4} & -7.2682 & 7.2682 & -16.1288 & 1.5924 \end{pmatrix} & \begin{pmatrix} \text{Be} & \text{E} & \text{T} & \text{V}_{ne} & \text{V}_{ee} \\ \text{WF1} & -13.5977 & 13.5977 & -29.5000 & 2.3047 \\ \text{WF2} & -13.6098 & 13.6098 & -29.4995 & 2.2799 \\ \text{WF3} & -13.6230 & 13.6230 & -29.4978 & 2.2519 \\ \text{WF4} & -13.6441 & 13.6441 & -29.5025 & 2.2144 \end{pmatrix} \end{matrix} \nonumber \] Except for a hiccup in the hydrogen anion results for WF4, these tables show that the improved agreement with experimental results (the lower total energy), is due to a reduction in electron-electron repulsion.