Uncertainty Product for a hydrogen-like atom,

In summary, to calculate the uncertainty product ΔrΔp for the 1s electron of a hydrogen-like atom with atomic number Z, you can use the wavefunction (Z/a0)^(3/2)*2exp(-Zr/a0) and equations such as E = (-13.6 eV)(Z^2/n^2), Δr = √(<r^2> - <r>^2), Δp = √(<p^2> - <p>^2), and <f> = ∫f(r)*P(r)dr. To find the uncertainty in momentum, you can use the formula <p2>=∫ψ*p2ψd3r
  • #1
dweeegs
12
1

Homework Statement



Calculate the uncertainty product ΔrΔp for the 1s electron of a hydrogen-like atom with atomic number Z. (Hint: Use <p> = 0 by symmetry and deduce <p^2> from the average kinetic energy)

Homework Equations



All I have is the wavefunction. For a 1s, it takes the form:

wavefunction = (Z/a0)^(3/2)*2exp(-Zr/a0)

Where Z is the atomic number, a0 is the Bohr radius

Other equations that I think I need include:

The energy for a hydrogen-like atom: E = (-13.6 eV)(Z^2/n^2)

Δr = √(<r^2> - <r>^2)
Δp = √(<p^2> - <p>^2)

<f> = ∫f(r)*P(r)dr

That integral is from 0 to ∞, and the <> are supposed to denote averages.

The Attempt at a Solution



I can find Δr no problem. Just use the above formula, where f(r) is r^2 for <r^2> and f(r) is r for <r>^2. The problem is finding the uncertainty in momentum. The above formula for momentum uncertainty should reduce down to

Δp = √(<p^2>)

And that's where I'm stuck; the hint isn't helping me much.

For another problem, I found the uncertainty in KE by finding the uncertainty in <U> for a hydrogen atom, and then used <KE> + <U> = <E>, where <E> = -13.6/n^2.

I'm not sure what the potential for a hydrogen-like atom is. Also, how can I get KE into momentum? I'm just confused on this part as a whole.

Any help is appreciated! Thank you!
 
Physics news on Phys.org
  • #2
you may probably calculate <p2> by using
<p2>=∫ψ*p2ψd3r,then apply first quantized form as p=-ih-∂/∂r and then you have to do some by part or a simple straightforward way without by part and done.On the other hand you might want to use
k.E.=p2/2m
 
  • #3
Am I too late?

It's no problem to calculate the expectation values of the position operators with a position-space wavefunction right? So it should be just as easy to calculate the expecation values of momentum operators with a momentum-space wavefunction!

If you find the momentum-space wavefunction, you are in the clear. My method completely ignores the hint though.
 

1. What is the uncertainty product for a hydrogen-like atom?

The uncertainty product for a hydrogen-like atom is a mathematical expression that describes the relationship between the uncertainty in position and the uncertainty in momentum of the electron in the atom. It is also known as the Heisenberg uncertainty principle for atoms.

2. How is the uncertainty product calculated for a hydrogen-like atom?

The uncertainty product for a hydrogen-like atom is calculated by multiplying the uncertainty in position (Δx) by the uncertainty in momentum (Δp) of the electron. The formula is given by ΔxΔp ≥ h/4π, where h is the Planck's constant.

3. What is the significance of the uncertainty product for a hydrogen-like atom?

The uncertainty product for a hydrogen-like atom is significant because it shows that the position and momentum of an electron in an atom cannot be simultaneously known with absolute certainty. This is a fundamental principle of quantum mechanics and has implications for our understanding of the behavior of particles at the atomic level.

4. How does the uncertainty product impact our understanding of the electron's behavior in a hydrogen-like atom?

The uncertainty product tells us that the electron's position and momentum cannot be precisely determined at the same time. This means that the electron's behavior in a hydrogen-like atom cannot be predicted with certainty, and there will always be a degree of uncertainty in our measurements and observations.

5. Can the uncertainty product for a hydrogen-like atom be applied to other atoms or particles?

Yes, the uncertainty product for a hydrogen-like atom can be applied to other atoms and particles. However, the specific values for each uncertainty may vary depending on the properties of the atom or particle in question. The uncertainty product is a fundamental principle of quantum mechanics and applies to all particles at the atomic level.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
14
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
Replies
1
Views
1K
Back
Top