Expectation value for electron in groundstate

In summary, the expectation value for r for an electron in the groundstate of a one-electron-atom is: (3/2)a_{0}/Z.
  • #1
Cp.L
6
0

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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  • #2
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
 
  • #3
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]
 
  • #4
Show your work.
 
  • #5
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]
 
  • #6
The integral should be from 0 to infinity. You also have a mistake somewhere else as your expression has units of 1/length.
 
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  • #7
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..
 
  • #8
How did you find C100?
 
  • #9
It was listed in my book, for the groundstate of an hydrogen atom so it should actually be correct.
 
  • #10
You need to go back and review spherical coordinates so you can set up the integrals correctly. In particular, you need to fix your limits and get the volume element right.
 
  • #11
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
 
  • #12
Hi, thanks alot. It really helped to use spherical coordinates :) got the correct answer now :)
 

FAQ: Expectation value for electron in groundstate

1. What is the expectation value for an electron in the ground state?

The expectation value for an electron in the ground state is the average value of a measurement that would be obtained if the electron's position or momentum were measured many times. It is a statistical concept that gives insight into the most likely position or momentum of an electron in its lowest energy state.

2. How is the expectation value for an electron in the ground state calculated?

The expectation value for an electron in the ground state is calculated by taking the integral of the wave function squared multiplied by the position or momentum operator. This integral represents the average value of the observable for the electron in its ground state.

3. What does the expectation value for an electron in the ground state tell us?

The expectation value for an electron in the ground state can tell us the most probable location or momentum of the electron, as well as the spread or uncertainty in these values. It also provides information about the energy of the electron in the ground state.

4. How does the expectation value for an electron in the ground state change with different potentials?

The expectation value for an electron in the ground state can change with different potentials, as the wave function and energy levels of the electron are affected by the potential. For example, in a stronger potential well, the expectation value for position will be closer to the center of the well, while in a weaker potential, it will be more spread out.

5. Why is the expectation value for an electron in the ground state important in quantum mechanics?

The expectation value for an electron in the ground state is important in quantum mechanics because it allows us to make predictions about the behavior of particles at the microscopic level. It also helps us understand the probabilistic nature of quantum systems and how they differ from classical systems.

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