Hydrogenoid atom | Quantum mechanics
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- Written by: Germán Fernández
- Category: Hydrogenoid atom | Quantum mechanics
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Atoms that only have one electron are called hydrogenoids. They are hydrogenoids atoms: H, He + , Li 2+ , Be 3+ .
Hydrogen atoms present a factorable wave function in a radial part and a spherical harmonic, which has the form:
$\Psi(r,\theta,\varphi)=R_{n,l}(r)Y_{l,m}(\theta,\varphi)$
where n,l,m are the quantum numbers that make the wave function acceptable.
The energy of a hydrogen atom is given by the expression:
$E_n=\frac{RhZ^2}{n^2}$
As can be seen, the energy depends exclusively on the principal quantum number, n.
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- Written by: Germán Fernández
- Category: Hydrogenoid atom | Quantum mechanics
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A central force is one that comes from a potential energy function with spherical symmetry, that is, a function that only depends on the distance to the origin of the particle: $V=V(r)$ Thus: $\left(\frac{ \partial V}{\partial\theta}\right)_{r,\varphi}=0 ;$ and $\left(\frac{\partial V}{\partial\varphi}\right)_{r,\ theta}=0 ;$
Let us now consider the quantum mechanics of a simple particle subjected to a central force:
\begin{equation} \hat{H}=\hat{T}+\hat{V}=-\frac{\hbar^2}{2m}\nabla^2+V(r) \end{equation}
Let's express $\nabla^2$ in spherical polar coordinates:
\begin{equation} \nabla^2=\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{1}{r^2}\cot\theta\frac{\partial}{\partial\theta}+\frac{1}{r^2 sin^2\theta}\frac {\partial^2}{\partial\varphi^2} \end{equation}
Remembering the expression of the operator $\hat{l}^2$:
\begin{equation}\label{ec3} \hat{l}^ 2=-\hbar^2\left(\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac{1}{sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right) \end{equation}
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- Written by: Germán Fernández
- Category: Hydrogenoid atom | Quantum mechanics
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It is a system of two particles separated by a distance d with no possibility of vibration $r=d$. The energy of the rotor is kinetic, therefore $V=0$. We are only interested in rotational energy.
Since r is constant, we can omit the factor $R(r)$ in the wave function, which will be given by a spherical harmonic. \begin{equation} \Psi =Y_{l,m}(\theta,\varphi) \end{equation} We set up the Schrödinger equation: \begin{equation} \left[-\frac{\hbar^2}{ 2\mu}\left(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right)+\frac{1 }{2\mu d^2}\hat{l}^2+V(r)\right]Y_{l,m}(\theta,\varphi)=EY_{l,m}(\theta,\varphi ) \end{equation} The derivatives with respect to $r$ are null, since the function does not depend on this variable.
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- Written by: Germán Fernández
- Category: Hydrogenoid atom | Quantum mechanics
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We set up the Schrödinger equation: \begin{equation}\label{ec1} \left[-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{r}\right ]\Psi=E\Psi \end{equation} Where $\Psi=R_{n,l}(r)Y_{l,m}(\theta,\varphi)$
Substituting the wave function into the Schödinger equation \begin{equation} -\frac{\hbar^2}{2\mu}\left(R''+\frac{2}{r}R'\right)+ \frac{l(l+1)\hbar^2}{2\mu r^2}R-\frac{Ze^2}{r}R=ER \label{ec2} \end{equation}
We simplify by multiplying all the terms of the equation (\ref{ec2}) by $-\frac{2\mu}{\hbar^2}$
\begin{equation} R''+\frac{2}{r}R'+\frac{2\mu ER}{\hbar^2}+\frac{2\mu Ze^2}{r\hbar^2}R-\frac{l(l+1)}{r^2}R= 0 \end{equation}
Putting $a=\frac{\hbar^2}{\mu e^2}$ into the equation (8) gives:
\begin{equation}\label{ec4} R''+\frac{2}{r}R'+\left[\frac{2E}{ae^2}+\frac{2Z}{ra}-\frac{l(l+1 )}{r^2}\right]R=0 \end{equation}
We solve the equation (\ref{ec4}) in two cases: