Calculating p^2/2m and -e^2/r for Hydrogen Atom in 3D

In summary, the conversation discusses a question related to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom in 3D. The individual attempted to calculate the averages using the wavefunction for the ground state, but ran into problems with an integral involving a term that seemed to explode at the origin/infinity. After realizing that the volume element provides an extra factor of r^2, they were able to successfully calculate the averages by multiplying by 4 pi after the integration in r.
  • #1
silimay
26
0

Homework Statement



I have a question on my quantum pset relating to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom (in 3D).

The Attempt at a Solution



I started out trying to calculate the averages with [tex]\psi[/tex] ... something like, for the ground state, [tex]\psi = \frac{e^{-2r/a_0}}{\sqrt{\pi * a_0^3}}[/tex].

But then I ran into problems (when I was trying to do the <-e^2/r>) when I came up with an integral involving a term [tex]\frac{e^{-2r/a_0}}{r}[/tex]. As far as I could see, this integral sort of seems to explode at the origin / at infinity. I was wondering, should I use just the radial wavefunction part, since it has an extra r factor that would make this integral possible? I was just confused basically...
 
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  • #2
silimay said:

Homework Statement



I have a question on my quantum pset relating to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom (in 3D).

The Attempt at a Solution



I started out trying to calculate the averages with [tex]\psi[/tex] ... something like, for the ground state, [tex]\psi = \frac{e^{-2r/a_0}}{\sqrt{\pi * a_0^3}}[/tex].

But then I ran into problems (when I was trying to do the <-e^2/r>) when I came up with an integral involving a term [tex]\frac{e^{-2r/a_0}}{r}[/tex]. As far as I could see, this integral sort of seems to explode at the origin / at infinity. I was wondering, should I use just the radial wavefunction part, since it has an extra r factor that would make this integral possible? I was just confused basically...

You are not giving the details of what you calculated so it's hard to answer but you should not have any problem because the volume element [tex] dV = r^2 \sin \theta ~dr d\theta d\phi [/tex] provides an extra factor of r^2. Did you take this into account?
 
  • #3
mmm, I didn't :) Thanks so much :)

I know it was smth silly I wasn't paying attention to ^_^

It's not dependent on theta or phi, so I should just multiply it by 4 pi after doing the integral in r, right?
 
  • #4
silimay said:
mmm, I didn't :) Thanks so much :)

I know it was smth silly I wasn't paying attention to ^_^

It's not dependent on theta or phi, so I should just multiply it by 4 pi after doing the integral in r, right?

That's correct. The integral over phi and theta of the angular part of the volume element gives 4 pi (you should do it once explicitly to see how it works out!)

Glad I could help.
 

FAQ: Calculating p^2/2m and -e^2/r for Hydrogen Atom in 3D

How do you calculate p^2/2m for a hydrogen atom in 3D?

In order to calculate p^2/2m for a hydrogen atom in 3D, you will need to use the formula p^2/2m = -h^2/2m * (∂^2/∂x^2 + ∂^2/∂y^2 + ∂^2/∂z^2), where h is Planck's constant and ∂ represents the partial derivative. This formula takes into account the momentum of the electron in all three dimensions, and results in the kinetic energy of the electron in the atom.

What is the significance of p^2/2m in the calculation for a hydrogen atom?

The expression p^2/2m represents the kinetic energy of the electron in the hydrogen atom. It is derived from the Schrödinger equation, which describes the behavior of quantum particles such as electrons. This calculation is important because it helps us understand the behavior and energy levels of electrons in atoms.

How do you calculate -e^2/r for a hydrogen atom in 3D?

To calculate -e^2/r for a hydrogen atom in 3D, you will need to use the formula -e^2/r = -ke^2/r, where k is Coulomb's constant and e is the charge of the electron. This formula takes into account the attractive force between the positively charged nucleus and the negatively charged electron, resulting in the potential energy of the electron in the atom.

What is the significance of -e^2/r in the calculation for a hydrogen atom?

The expression -e^2/r represents the potential energy of the electron in the hydrogen atom. It is derived from the Coulomb potential, which describes the electrostatic interaction between charged particles. This calculation is important because it helps us understand the stability and energy levels of the electron in the atom.

How do you use p^2/2m and -e^2/r to calculate the total energy of a hydrogen atom in 3D?

To calculate the total energy of a hydrogen atom in 3D, you will need to use the formula E = p^2/2m - e^2/r. This formula takes into account both the kinetic and potential energies of the electron in the atom, resulting in the overall energy of the system. By calculating the energy levels of the electron in the atom, we can better understand its behavior and properties.

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