Ground state energy of hydrogen atom

[Bosonichadron]

Hi everyone, This question is from my problem set this week in my Phys 371 class. Any help, hints or ideas would be very much appreciated!

"Use the Heisenberg Uncertainty Principle to estimate the ground state energy in the hydrogen atom. Since the wave function that solves this problem is not a Gaussian, it will work best if you use \sigma_{r}\sigma_{p}=\hbar."

Where \sigma_{r} is the standard deviation of the radius centered at the nucleus and \sigma_{p} is the standard deviation of the momentum of the electron.

What I tried so far is to get the momentum in terms of the kinetic energy p=sqrt(2m(E-V)) [where V is the potential energy and E-V is the kinetic] and then put V in terms of r, since it would just be the coulomb potential energy...the trouble is that the algebra is devastatingly complicated and rather tedious when I try to solve for E--so it seems like there should be an easier way. Also, seems dubious to have E in terms of r without knowing what r is.

 

[Marty]

Hi everyone, This question is from my problem set this week in my Phys 371 class. Any help, hints or ideas would be very much appreciated!

"Use the Heisenberg Uncertainty Principle to estimate the ground state energy in the hydrogen atom. Since the wave function that solves this problem is not a Gaussian, it will work best if you use \sigma_{r}\sigma_{p}=\hbar."

Where \sigma_{r} is the standard deviation of the radius centered at the nucleus and \sigma_{p} is the standard deviation of the momentum of the electron.

What I tried so far is to get the momentum in terms of the kinetic energy p=sqrt(2m(E-V)) [where V is the potential energy and E-V is the kinetic] and then put V in terms of r, since it would just be the coulomb potential energy...the trouble is that the algebra is devastatingly complicated and rather tedious when I try to solve for E--so it seems like there should be an easier way. Also, seems dubious to have E in terms of r without knowing what r is.

I don't know if this is cheating, but there is a fact of astronomy which you can verify pretty easily whereby the kinetic energy of a satellite in orbit is numerically equal to the gravitational potential energy, but opposite in sign. If you allow the same thing to apply to the hydrogen atom, you can simplify some of your algebra. Your term in E-V just becomes equal to V (but opposite in sign: that is, the KE is positive and the PE is negative.)

 

[h0dgey84bc]

Well if you assume uncertainty in each component of position is of the order of magnitude of the Bohr radius a_0. Then (approx) \Delta P_x = \frac{\hbar}{a_0}. If average components of momentum have p_x= \Delta p_x , p_y=\Delta p_y etc. Then the average KE is \frac{p^2}{2m} =3 \frac {\hbar^2}{8ma^2}. Which comes out at around 10eV. So not bad for such a crappy method of guessing.