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==Modeling by quantum-mechanical oscillator== The [[van der Waals force|van der Waals]] binding energy can be analyzed by another simple physical picture: modeling the motion of an electron around its nucleus by a three-dimensional simple [[harmonic oscillator]] with a potential energy ''V<sub>a</sub>'':{{Clarify|reason=Provide a geometric image representing the equitation|date=August 2021}} :<math>V_a = \frac{m_e}{2}{\omega^2}(x^2+y^2+z^2),</math> where ''m<sub>e</sub>'' and ''ω'' are the mass and vibrational frequency of the electron, respectively. As this atom approaches the surface of a metal and forms adsorption, this potential energy ''V<sub>a</sub>'' will be modified due to the image charges by additional potential terms which are quadratic in the displacements: :<math>V_a = \frac{m_e}{2}{\omega^2}(x^2+y^2+z^2)-{e^2\over 16\pi\varepsilon_0 Z^3}\left(\frac{x^2+y^2}{2}+z^2\right)+\ldots</math> (from the Taylor expansion above.) Assuming :<math> m_e \omega^2\gg{e^2\over 16\pi\varepsilon_0 Z^3},</math> the potential is well approximated as :<math>V_a \sim \frac{m_e}{2}{\omega_1^2}(x^2+y^2)+\frac{m_e}{2}{\omega_2^2}z^2</math>, where :<math> \begin{align} \omega_1 &= \omega - {e^2\over 32\pi\varepsilon_0 m_e\omega Z^3},\\ \omega_2 &= \omega - {e^2\over 16\pi\varepsilon_0 m_e\omega Z^3}. \end{align} </math> If one assumes that the electron is in the ground state, then the van der Waals binding energy is essentially the change of the zero-point energy: :<math>V_v = \frac{\hbar}{2}(2\omega_1+\omega_2-3\omega)= - {\hbar e^2\over 16\pi\varepsilon_0 m_e\omega Z^3}.</math> This expression also shows the nature of the ''Z''<sup>−3</sup> dependence of the van der Waals interaction. Furthermore, by introducing the atomic [[polarizability]], :<math> \alpha= \frac {e^2} {m_e\omega^2},</math> the van der Waals potential can be further simplified: :<math>V_v = - {\hbar \alpha \omega\over 16\pi\varepsilon_0 Z^3}= -\frac{C_v}{Z^3},</math> where :<math>C_v = {\hbar \alpha \omega\over 16\pi\varepsilon_0},</math> is the van der Waals constant which is related to the atomic polarizability. Also, by expressing the fourth-order correction in the Taylor expansion above as (''aC<sub>v</sub>Z''<sub>0</sub>) / (Z<sup>4</sup>), where ''a'' is some constant, we can define ''Z''<sub>0</sub> as the position of the ''dynamical image plane'' and obtain {| class="wikitable floatright" |+Table 1. The van der Waals constant ''C<sub>v</sub>'' and the position of the dynamical image plane ''Z''<sub>0</sub> for various rare gases atoms adsorbed on noble metal surfaces obtained by the jellium model. Note that ''C<sub>v</sub>'' is in eV/Γ <sup>3</sup> and ''Z''<sub>0</sub> in Γ . ! rowspan="2" | ! colspan="2" | He !colspan="2" | Ne !colspan="2" | Ar !colspan="2" | Kr !colspan="2" | Xe |- !''C<sub>v</sub>'' !''Z''<sub>0</sub> !''C<sub>v</sub>'' !''Z''<sub>0</sub> !''C<sub>v</sub>'' !''Z''<sub>0</sub> !''C<sub>v</sub>'' !''Z''<sub>0</sub> !''C<sub>v</sub>'' !''Z''<sub>0</sub> |- !Cu |0.225 |0.22 |0.452 |0.21 |1.501 |0.26 |2.11 |0.27 |3.085 |0.29 |- !Ag |0.249 |0.2 |0.502 |0.19 |1.623 |0.24 |2.263 |0.25 |3.277 |0.27 |- !Au |0.274 |0.16 |0.554 |0.15 |1.768 |0.19 |2.455 |0.2 |3.533 |0.22 |} :<math>V_v = - \frac{C_v}{(Z-Z_0)^3}+O\left(\frac{1}{Z^5}\right).</math> The origin of ''Z''<sub>0</sub> comes from the spilling of the electron wavefunction out of the surface. As a result, the position of the image plane representing the reference for the space coordinate is different from the substrate surface itself and modified by ''Z''<sub>0</sub>. Table 1 shows the [[jellium]] model calculation for van der Waals constant ''C<sub>v</sub>'' and dynamical image plane ''Z''<sub>0</sub> of rare gas atoms on various metal surfaces. The increasing of ''C<sub>v</sub>'' from He to Xe for all metal substrates is caused by the larger atomic [[polarizability]] of the heavier rare gas atoms. For the position of the dynamical image plane, it decreases with increasing dielectric function and is typically on the order of 0.2 Γ .
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