How to Find Frequency of a Hydrogen Atom if You Dont Know N Final

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Solving the Schrödinger equation for hydrogen-like atoms

Consider the Schrödinger (time contained) Wave Equation

$\begin{displaymath} -\frac{\hbar ^2}{2m}\nabla ^2 \psi + U\psi = E\psi \end{displaymath}$

(1)

which is expanded as

$\begin{displaymath} \left( \frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}... ...^2}{\partial z^2} \right)\psi + \frac{2m}{\hbar^2}(E-U)\psi =0 \end{displaymath}$

(ii)

in Cartesian co-ordinates.

When applied to the hydrogen atom, the wave function should describe the behaviour of both the nucleus and the electron, $\psi \Rightarrow \psi ({\bf r}_n,{\bf r}_e)$ . This means we have a two torso problem, which is very difficult to solve. We can fortunately convert this ii-body problem to an constructive one-body trouble by transforming from the Laboratory System to the Centre of Mass System.

$\begin{displaymath} ({\bf r}_n,{\bf r}_e) \Rightarrow ({\bf R},{\bf r}). \end{displaymath}$

(iii)

In this system, the wave function is separable, $\psi \Rightarrow \psi ({\bf R})\psi ({\bf r})$ . (Can you call back why ?) In addition, nosotros note that the CofM motion is constant in the absence of external forces. Thus we need simply solve the wave equation for the behaviour of $\psi({\bf r})$ . Then for studying hydrogen-like atoms themselves, we demand but consider the relative motion of the electron with respect to the nucleus. Notation that in this instance the appropriate mass to employ in the moving ridge equation will be the reduced mass of the electron, $m_e \Rightarrow m_r = m_n m_e/(m_n+m_e)$ . Here ${\bf R},{\bf r}$ are the position vectors to the CofM and the electron w.r.t. the nucleus respectively and the significant of the other symbols are obvious.

The appropriate potential is of course the simple radial electrostatic potential for a point accuse nucleus of charge . (Note that if nosotros are not dealing with hydrogen, and so nosotros are dealing with hydrogen-like atoms which are fully stripped of all just one electron.)

$\begin{displaymath} U(x,y,x) = U({\bf r}) = U(r) =-\frac{Ze^2}{4\pi \epsilon_0 r} \end{displaymath}$

(4)

has spherical symmetry. One could write $r=\sqrt{x^2+y^2+z^2}$ and solve the wave equation in Cartesian co-ordinates. This would piece of work just information technology would be very tedious, as the mathematics does non brandish the symmetry of the physics. Accordingly we rather exploit the spherical symmetry of the potential, and perform a co-ordinate transformation from Cartesian Co-ordinates (efficient for rectangle shapes) to Spherical Polar Co-ordinates (efficient for spherical shapes)

$\begin{displaymath} (x,y,z) \longrightarrow (r,\theta,\phi). \end{displaymath}$

(5)

These new co-ordinates are defined in figure i. Under this co-ordinate transformation, the wave equation takes the form,

$\begin{displaymath} \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\par... ...rtial^2\psi}{\partial \phi^2} + \frac{2m}{\hbar^2}(E-U)\psi =0 \end{displaymath}$

(6)

Exercise ane Writing the double differential operator in spherical polar co-ordinates is non trivial. Bank check Electromagnetic Fields and Waves by Lorrain and Carson, Chapter 1, if you require insight into the last step. You are not expected to be able to do this transformation. Notwithstanding, make certain, using a sketch, that you tin can prove how an infinitesimal volume element behaves under this transformation.

$\begin{displaymath} dv = dx\,dy\,dz = r^2 \sin\theta\,dr\,d\theta\,d\phi = d^3r \end{displaymath}$

(vii)

**Figure 1:** Cartesian and Spherical Polar co-ordinates
$\includegraphics[width=0.5\textwidth]{Sph_Polar.eps}$

Practise 2
Write each of the variables $(r,\theta,\phi)$ in terms of the variables , too perform the inverse mapping. You may use figure 1. Note that the potential is radial, $U({\bf r})=U(r,\theta,\phi)=U(r)$ , which means it depends only on , and not on $\theta$ or $\phi$ .

The wave function necessarily is separable into radial, polar and azimuthal factors nether a radial potential equally follows : $\psi (r,\theta ,\phi) = R(r)\Theta(\theta)\Phi(\phi)$ . (One time once again, tin can you think why ?) Substituting the above expression and the potential into the spherical polar representation of the wave equation, nosotros notice, after some manipulation,

$\begin{displaymath} \frac{\sin^2 \theta}{R}\frac{\partial}{\partial r}\left( {r^... ...E\right) =-\frac{1}{\Phi}\frac{\partial^2\Phi}{\partial\phi^2} \end{displaymath}$

(8)

Exercise 3
Ensure you can achieve the final consequence with your own pencil and paper. Doing these exercises and tutorials properly helps to make you familiar with quantum mechanics, via your fingertips, into the marrow of your bones. You may use your textbooks the offset time y'all practise this.

This equation has on the left functions of $(r,\theta)$ merely, and on the right a function of $(\phi)$ only. Accordingly, it can only be satisfied for all values of the independent variables by requiring that both sides are equal to the same constant value. Hence it follows the the left hand side, which is called the azimuthal equation, is equal to a constant, which is called the azimuthal abiding. Nosotros give the azimuthal abiding the symbol , for reasons which will get articulate with retrospect.

$\begin{displaymath} -\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = \mbox{const(azimuthal)} = m_l^2 \end{displaymath}$

(9)

The solution for the azimuthal equation (by the ansatz method) is

$\begin{displaymath} \Phi_{m_l}(\phi)=Ae^{im_l\phi} \end{displaymath}$

(10)

Exercise 4
Bank check this with your own pencil and newspaper....

Imposing the boundary conditions, nosotros must have $\Phi(\phi)$ single valued, and $\Phi(\phi) = \Phi(\phi+2\pi)$ . Therefore, we require the constant to be quantised as follows.

$\begin{displaymath} m_l = 0, \pm 1, \pm 2, \pm 3, \cdots \end{displaymath}$

(11)

Nosotros phone call the magnetic breakthrough number. You may imagine that the quantity $\frac{d\Phi}{d\phi}$ measures the change of the azimuthal office of the wave part with modify of the azimuthal according. Thus this quantity tin can sense differences nether a rotation about the -axis. Intuitively, if the wave function of the electron changed under such a rotation, i would be able to discern it, and classically, a rotating accuse has a magnetic moment. This is a heuristic justification of the appellation of ``magnetic quantum number'' for .

Clearly, non only is the separation constant quantised, simply also the azimuthal wave function becomes part of a family of wave functions labelled by the breakthrough number .

Congratulations !

You have but seen that the quantisation of the magnetic quantum number arises naturally from the condition that the wave part must be single valued and satisfy its boundary conditions. The imposition of boundary conditions also lead to quantisation of the wave function in the previous examples we have seen (particle in a box). This is just a mathematical way of saying nosotros have successfully trapped a moving ridge role in an attractive potential. The value of the prefactor will be set subsequently via a normalisation condition. Thus fortified, we proceed to the polar wave function. Substituting the abiding, (known every bit a separation constant), back into the wave equation above and re-arranging terms

$\begin{displaymath} \frac{1}{R}\frac{\partial}{\partial r}\left( {r^2}\frac{\par... ...} \left(\sin\theta\frac{\partial\Theta}{\partial\theta}\right) \end{displaymath}$

(13)

Once more, we have an equation which has on the left functions of

merely, and on the correct functions of $\theta$ but. Accordingly, information technology tin only be satisfied for all values of the independent variables by requiring that both sides are equal to the same constant value. Hence it follows,

We have achieved so far separated equations for the last two wave functions, viz. the radial wave equation for and the polar moving ridge equation for $\Theta(\theta)$ . The solution of these two equations is beyond the telescopic of this course. Residual assured, it gain every bit in the case for the azimuthal wave function. That is, imposing the boundary weather condition causes the separation constant to become quantised and also the radial moving ridge function and the polar wave function to go part of a family labelled by the advisable quantum number.

(The bending dependent wave functions are known as the Spherical Harmonic Functions, and the radial wave functions as Laguerre polynomials. These solutions are tabulated beneath in figure 2.)

**Figure ii:** Normalised wave functions of the hydrogen cantlet for and 3.

Hence we find, on solving the polar wave equation for $\Theta(\theta)$ that

$\begin{displaymath} \mbox{const(polar)} = l(l+1) \mbox{\ \ where \ \ } l \ge \vert m_l\vert \mbox{ is an integer} \end{displaymath}$

(16)

so that

$\begin{displaymath} m_l = 0, \pm 1, \pm 2, \pm 3, \cdots ,\pm l \end{displaymath}$

(17)

We phone call

the orbital quantum number. In a subsequently department, we will testify the human relationship of this quantum number to orbital angular momentum. This is then the heuristic justification of its appellation.

Exercise five
What do y'all think the boundary conditions are for the polar wave equation ?

In the introductory grade on Quantum Mechanics, we saw that was also a separation abiding. It arose when nosotros separated the time and space parts of the Time dependent wave equation to get in at the Time independent wave equation, which we accept presented at the top of this section. So it will come equally no surprise now to find that becomes quantised on requiring that the solutions of the radial wave equation above besides obey boundary weather. In the case of the radial moving ridge equation, we obviously require that is finite and $R(r=\infty) = 0$ . We get ,

$\begin{displaymath} E_n = -\frac{mZ^2e^4}{32\pi^2\epsilon^2_0\hbar^2} \left( \frac{1}{n^2} \right) = \frac{E_1}{n^2}, \ \ \ n=1,2,3,\cdots \end{displaymath}$

(18)

with the ancillary status that

$\begin{displaymath} n \ge l+1 \mbox{ or } l = 0, 1, 2, \cdots ,(n-1) \end{displaymath}$

(19)

We phone call

the principal quantum number.

Exercise 6
This is the same expression as in the Bohr model. Muse on what this correspondence with the Bohr model implies.

Exercise 7
How would these results change if we were dealing with positronium and not hydrogen ?

Nosotros may summarise our results so far :
The Schrödinger Wave Equation for hydrogen like atoms in three dimensions is best treated in spherical polar co-ordinates

$\begin{displaymath} (x,y,z) \longrightarrow (r,\theta,\phi) \end{displaymath}$

(20)

considering the Coulomb potential for this example

$\begin{displaymath} U(x,y,x) = U({\bf r}) = U(r) =-\frac{Ze^2}{4\pi \epsilon_0 r} \end{displaymath}$

(21)

is spherically symmetric. Consequently, the wave part will be separable

$\begin{displaymath} \psi ({\bf r}) = R(r)\Theta(\theta)\Phi(\phi) \end{displaymath}$

(22)

and the Schrödinger Wave Equation reduces to the 3 equations

$\begin{displaymath} \frac{d^2\Phi}{d\phi^2} - m_l^2 \Phi = 0 , \end{displaymath}$

(23)

$\begin{displaymath} \frac{\hbar^2}{2m} \frac{1}{r^2}\frac{d}{dr}\left( {r^2}\fra... ..._0 } \frac{1}{r} -\frac{l(l+1)\hbar^2}{2mr^2} -E \right]R = 0 \end{displaymath}$

(24)

and

$\begin{displaymath} \frac{1}{\sin \theta} \frac{d}{d\theta} \left(\sin\theta\fra... ...eft [ l(l+1) - \frac{m_l^2}{\sin^2 \theta} \right]\Theta = 0. \end{displaymath}$

(25)

Solving these equations subject area to the appropriate boundary atmospheric condition leads to the three sets of breakthrough numbers :

$\begin{eqnarray*} \mbox{ Principal Quantum Number \ \ } & n = 1,2,3, \cdots \\ ... ...antum Number \ \ } & m_l = 0, \pm 1, \pm 2, \pm 3, \cdots ,\pm l \end{eqnarray*}$

In this procedure, nosotros also detect a family of wave functions, labelled by the breakthrough numbers :

$\begin{displaymath} \psi_{nlm_l} ({\bf r}) = R_{nl}(r)\Theta_{lm_l}(\theta)\Phi_{m_l}(\phi). \end{displaymath}$

(26)

We can call back of the set of quantum numbers

every bit identifying a wave function for a particular state $\psi_{nlm_l} ({\bf r})$ . It is typical that quantum numbers appear naturally when quantum particles are trapped in a detail region of space by an bonny potential. A choice of the lowest energy moving ridge functions take been collected in the table higher up. These moving ridge functions are normalised and so that the probability density for finding an electron in a particular state

represented past $\vert\psi_{n,l,m_l} (r,\theta ,\phi)\vert^2$ is unity when integrated over all space.

$\begin{displaymath} \int^{\infty}_0 \int^{\pi}_0 \int^{2\pi}_0 \vert\psi_{n,l,m_l} (r,\theta ,\phi)\vert^2 \, dr.rd\theta .r\sin\theta d\phi = 1 \end{displaymath}$

(27)

Ascertainment
Schrödinger sure deserved the Nobel Prize !

Instance
Suppose we want to verify the free energy of the ground state wave role of the hydrogen atom, and . Nosotros annotation that information technology is only the radial moving ridge equation which contains , the energy of the state. The appropriate radial wave function is

$\begin{displaymath} R(r) = \frac{2}{a_0^{3/2}}e^{-r/a_0}. \end{displaymath}$

(28)

We can substitute this wave part into the radial wave equation

$\begin{displaymath} \frac{\hbar^2}{2m} \frac{1}{r^2}\frac{d}{dr}\left( {r^2}\fra... ...2} +E \right] \left(\frac{2}{a_0^{3/2}}e^{-r/a_0} \right) = 0. \end{displaymath}$

(29)

Using

and

, we simplify to find

$\begin{displaymath} \left [ \left( \frac{2}{a_0^{7/2}} + \frac{4mE_1}{\hbar^2a_0... ...\frac{4}{a_0^{5/2}} \right)\frac{1}{r} \right]e^{-r/a_0} =0 . \end{displaymath}$

(thirty)

Each parenthesis must equal nought for the unabridged equation to equal nought. We therefore find an expression for

, the Bohr radius,

$\begin{displaymath} a_0 = \frac{4\pi\epsilon_0\hbar}{me^2} = r_1 \end{displaymath}$

(31)

and the ground state energy

$\begin{displaymath} E_1 = -\frac{\hbar^2}{2ma_0^2} = -\frac{me^2}{32\pi^2\epsilon_0^2\hbar^2} \end{displaymath}$

(32)

Next: Quantum numbers Up: Quantum Mechanics of Atoms Previous: A full Quantum Mechanical

Simon Connell 2004-10-04

steckerswee1941.blogspot.com

Source: http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-03/lecture_03/node4.html

How to Find Frequency of a Hydrogen Atom if You Dont Know N Final

Solving the Schrödinger equation for hydrogen-like atoms

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