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Topic: Energy of Hydrogen-like atoms using Shcordinger's formula  (Read 1547 times)

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Offline sixstorm1

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Energy of Hydrogen-like atoms using Shcordinger's formula
« on: October 11, 2016, 10:14:48 PM »
Hello,

Really need help for an homework, here is what I need to do:

For Hydrogen-like atoms, using Schrodinger's formula: Eψnlm=-(h²/2m)d²Eψnlm/dr²+VEψnlm (where V=-ke²/r)
a) Derive a formula for E=1,0,0 , E=2,0,0 and E=3,0,0 (where E=n,l,m) and show the second derivative for each wave function.
b) Calculate the value for each in Joules.

Please advise on how to do this. I have spent the last day looking at various resources on the topic (Schrödinger equation for the hydrogen and hydrogen-like atoms, Schrodinger equation in spherical coordinates, etc), at these places:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sch3d.html#c2
http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_8/node5.html
https://en.wikipedia.org/wiki/Hydrogen-like_atom
http://onlinelibrary.wiley.com/doi/10.1002/9780470747414.app9/pdf
http://en.citizendium.org/wiki/Hydrogen-like_atom

I understand the wave functions, and atomic orbitals and probability. But this above looks very unclear to me. Where do I start? Do I need to use spherical coordinates? In the given equation, ψnlm cancels out on each side of the equation, this makes no sense. Why? What do I need to derive exactly? Thank you very much
« Last Edit: October 11, 2016, 11:00:56 PM by sixstorm1 »

Offline Borek

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Re: Energy of Hydrogen-like atoms using Shcordinger's formula
« Reply #1 on: October 12, 2016, 03:26:41 AM »
Do I need to use spherical coordinates?

Equation already uses spherical coordinates. The only variable shown is r, which is a distance from the nucleus (actually from the mass center of the atom). Equation as listed looks to be

[tex]E\Psi_{nlm}=-\frac{\hbar^2}{2\mu}\frac{d^2 E \Psi_{nlm}}{dr^2} + VE\Psi_{nlm}[/tex]

which (as far as I can tell, quantum chemistry is not my forte) is not correct. Without some of the Es listed it would be probably correct only when you ignore angular part of the function. Could make sense, as you are asked to deal with l,m=0 cases only.

Quote
In the given equation, ψnlm cancels out on each side of the equation, this makes no sense.

It doesn't cancel out, second derivative of the function is not the same thing as the function itself.

Without knowing the context of the course it is hard to say what you are expected to do. Perhaps you need to solve the differential equation as given.
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Offline sixstorm1

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Re: Energy of Hydrogen-like atoms using Shcordinger's formula
« Reply #2 on: October 12, 2016, 09:27:41 AM »
Thank you for your reply. I have found out more about this.

He wants us to double-derive the normalized wave functions given here: http://plato.mercyhurst.edu/chemistry/kjircitano/InorgStudysheets/InorgWaveFunction.pdf . The question was very unclear.

When I solve the whole thing, there is just one small problem left, the double derivative of ψ1,0,0 is positive (https://www.wolframalpha.com/input/?i=derive(derive(1%2F(sqr(pi)*a%5E(3%2F2))*exp(-r%2Fa),r),r)). When I substitute this result into the given equation of my post above and solve it (https://www.wolframalpha.com/input/?i=solve+z*(1%2F(sqr(pi)*a%5E(3%2F2))*exp(-r%2Fa))%3D-h%5E2%2F(2*m)*exp(-r%2Fa)%2F(a%5E(7%2F2)*sqr(pi))%2B(-(k*c%5E2)%2Fr)*(1%2F(sqr(pi)*a%5E(3%2F2))*exp(-r%2Fa))+for+z), the Kinetic energy part of the equation is negative. The potential energy part of the equation being also negative, this gives a false result. When I magically make one negation sign disappear, then the answer is correct (-2.179*10^-18). But magic is not accepted as a solution. Any idea where the culprit might be?

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