Expectation value for electron in groundstate

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Cp.L
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Homework Statement


Show that the expectation value for r for an electron in the groundstate of a one-electron-atom is:
<r>=(3/2)a[itex]_{0}[/itex]/Z



Homework Equations


Expectationvalue:
<f(x)>=∫[itex]\psi[/itex]*f(x)[itex]\psi[/itex]dx, -∞<x>∞

[itex]\psi[/itex][itex]_{100}[/itex]=C[itex]_{100}[/itex] exp(-Zr/a[itex]_{0}[/itex]), [itex]a_{o}\ =\ 0.5291\ \times\ 10^{-10}m , h\ =\ 6.626\ \times\ 10^{-34}\ J\ s[/itex]


The Attempt at a Solution


Im stuck..
 
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Cp.L said:

Homework Statement


Show that the expectation value for r for an electron in the groundstate of a one-electron-atom is:
<r>=(3/2)a[itex]_{0}[/itex]/Z



Homework Equations


Expectationvalue:
<f(x)>=∫[itex]\psi[/itex]*f(x)[itex]\psi[/itex]dx, -∞<x>∞

[itex]\psi[/itex][itex]_{100}[/itex]=C[itex]_{100}[/itex] exp(-Zr/a[itex]_{0}[/itex]), [itex]a_{o}\ =\ 0.5291\ \times\ 10^{-10}m , h\ =\ 6.626\ \times\ 10^{-34}\ J\ s[/itex]


The Attempt at a Solution


Im stuck..

What is stopping you from doing the integral? Of course, first you must determine the correct value of C100; do you know how to do that?

RGV
 
Hi, yes i used that C[itex]_{100}[/itex] =(1/[itex]\sqrt{\pi}) (\frac{z}{a_{0}}[/itex])[itex]^{3/2}[/itex]

Then i do the integral, but i must be doing it wrong cause i end up with e[itex]^{\frac{-2zr}{a_{0}}}[/itex] in the answer and also a problem of e[itex]^{∞}[/itex]
 
Ok, i found a mistake, but still don't get it right.
I write:

<r>=[itex]\frac{1}{\pi}[/itex][itex]\frac{z_{0}^{3}}{a_{0}^{3}}[/itex]∫e[itex]^{\frac{-2Zr}{a_{0}}}[/itex]r dr

Where the limits are from -∞ to ∞

Integrating i let U= r, du=1, dv= e[itex]^{\frac{-2Zr}{a_{0}}}[/itex], V=[itex]\frac{a_{0}}{-2Z}[/itex]e[itex]^{\frac{-2Zr}{a_{0}}}[/itex]

this integration leaves me with e[itex]^{∞}[/itex], or is there some trick for e[itex]^{-∞}[/itex] - e[itex]^{∞}[/itex]
 
Oh yes that's logical since its the radius, thanks. Yes and the integration by parts should be
u=r
du=dr
dv=e[itex]^{\frac{-2Zr}{a_{0}}}[/itex]
v=[itex]_{}\frac{a_{0}}{-2Z}[/itex]e[itex]^{\frac{-2Zr}{a_{0}}}[/itex] dr

integrating [itex]\frac{z^{3}}{\pi a_{0}}[/itex]∫e[itex]^{\frac{-2Zr}{a_{0}}}[/itex] r dr from 0 to ∞ with these vaues i get
<r>= -[itex]\frac{Za_{0}}{4\pi}[/itex]

I wonder if my C[itex]_{100}[/itex] might be wrong..
 
It was listed in my book, for the groundstate of an hydrogen atom so it should actually be correct.
 
Cp.L said:
It was listed in my book, for the groundstate of an hydrogen atom so it should actually be correct.

You should verify for yourself the value given in the book. If you cannot get the book's value, that is a signal that you may be doing something wrong---or possibly that the book made an error---but you should go with the first hypothesis unless you have overwhelming evidence to the contrary.

Once you are able to get the correct value of C100 you will be in a better position to find <r>.

RGV
 
Hi, thanks a lot. It really helped to use spherical coordinates :) got the correct answer now :)