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Perturbation of a Hydrogen atom

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose there is a deviation from Coulomb's law at very small distances, with the mutual Coulomb potential energy between an electron and a proton being given by:
    $$V_{mod}(r)= \begin{cases} - \frac {e^2} {4 \pi \varepsilon_0} \frac {b} {r^2} & \text {for } 0 \lt r \leq b \\ - \frac {e^2} {4 \pi \varepsilon_0} \frac {1} {r} & \text {for } r \gt b \end{cases}$$
    where ##e## is the magnitude of the electon charge, ##\varepsilon_0## is the permittivity of free space, ##r## is the electron-proton seperation and ##b## is a constant length that is small compared to the Bohr radius but large compared to the radius of a proton. Throughout this question, the perturbed systen, with ##V(r)## replaced by ##V_{mod}(r)##, will be called the modified Coulomb model.

    a) Specify the perturbation for the modified Coulomb model of a hydrogen atom relative to the unperturbed Coulomb model.

    b) Use this perturbation to calculate the first-order correction, ##E_1^{(1)}## to the fround-state energy of a hydrogen atom in the modified Coulomb model, givesn that the fround-state energy eigenfunction for the unperturbed Coulomb model is:
    $$\psi_{1,0,0} \left( r,\theta,\phi \right) = \left( \frac {1} {\pi a_0^3} \right)^{1/2} e^{-r/a_0} $$
    c) Show that your answer to part (b) can be approximated by
    $$E_1^{(1)} \approx - \frac {4b^2} {a_0^2} E_R$$ where ##E_R = {e^2} / 8 \pi \varepsilon_0 a_0## is the Rydberg energy. Hence deduce the largest value of ##b## that would be consistent with the fact that the ground-state energy of a hydrogen atom agrees with the predictions of the Coulomb model to one part in a thousand. Express your answer as a numerical multiple of ##a_o##.
    2. Relevant equations
    $$\int_0^x e^{-u} du = 1-e^{-x}$$ $$\int_0^x u e^{-u} du = 1-e^{-x}-xe^{-x}$$ for ##x \ll 1##,$$e{-x}=1-x+ \frac {x^2} {2} $$

    3. The attempt at a solution
    a)
    ##\delta \hat {\mathbf H}= - \frac {e^2} {4 \pi \varepsilon_0} \left( \frac {b} {r^2} - \frac 1 r \right)##

    b)
    ##E_1^{(1)}= - \frac {e^2} {\pi \varepsilon_0 a_0^3} \left( \frac {a_0b} {2} \left(1-e^{-2b/a_0} \right) - \frac {a_0^2} {4} \left(1-e^{-2b/a_0}-\frac {2b} {a_0} e^{-2b/a_0} \right) \right) ##

    c)
    This is where I'm having problems. I can get ##E_1^{(1)} \approx - \frac {4b^2} {a_0^2} E_R## by setting ##e^{-2b/a_0}=1## since ##b \ll a_0## and then integrate in the same way I did to get to answer (b), but should I be using my answer to part (b) to show ##E_1^{(1)} \approx - \frac {4b^2} {a_0^2} E_R## because I can't make it do that.

    Also, I'm not sure how to proceed from here to deduce the largest value of ##b##, and I'm a bit unclear to what the question means by "agrees with the predictions of the Coulomb model to one part in a thousand".

    Can anyone offer and advice with this please.
     
  2. jcsd
  3. Mar 19, 2017 #2

    vela

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    I got a different result for (b). Can you show us your calculations?
     
  4. Mar 20, 2017 #3
    This is how I got to (b):
    $$\begin{align} E_1^{(1)} & = \int_0^\infty R_{nl}^*(r) \delta \hat {\mathbf H} R_{nl}(r) r^2 dr \nonumber \\ & = \int_0^b \left( \frac {1} {a_0} \right)^{3/2} 2e^{-r/a_0} \left( - \frac {e^2} {4 \pi \varepsilon_0} \right) \left( \frac {b} {r^2} - \frac 1 r \right) \left( \frac {1} {a_0} \right)^{3/2} 2e^{-r/a_0} r^2 dr \nonumber \\ & = \int_0^b \left( - \frac {e^2} {4 \pi \varepsilon_0} \right) \left( \frac {b} {r^2} - \frac 1 r \right) \frac {4} {a_0^3} e^{-2r/a_0} r^2 dr \nonumber \\ & = - \frac {e^2} {\pi \varepsilon_0 a_0^3} \int_0^b \left( \frac {b} {r^2} - \frac 1 r \right) e^{-2r/a_0} r^2 dr \nonumber \\ & = - \frac {e^2} {\pi \varepsilon_0 a_0^3} \int_0^b \left( b - r \right) e^{-2r/a_0} dr \nonumber \\ & = - \frac {e^2} {\pi \varepsilon_0 a_0^3} \left( \int_0^b b e^{-2r/a_0} dr - \int_0^b r e^{-2r/a_0} dr \right) \nonumber \end{align} $$
    Then I used the given integrals to get to:
    $$\int_0^b b e^{-2r/a_0} dr = \frac {a_0 b} {2} \int_0^{2b/a_0} e^{-u} du = \frac {a_0 b} {2} \left( 1 - e^{-2b/a_0} \right) $$
    And
    $$\int_0^b r e^{-2r/a_0} dr = \frac {a_0^2} {4} \int_0^{2b/a_0} u e^{-u} du = \frac {a_0^2} {4} \left( 1 - e^{-2b/a_0} - \frac {2b} {a_0}e^{-2b/a_0} \right) $$
    Then I substituted back in to get my answer for (b). Have I messed this up somewhere?
     
  5. Mar 20, 2017 #4

    vela

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    Oops, my mistake, I simplified your answer incorrectly. You have some terms that will cancel out, and you should end up with
    $$\frac{k e^2}{a}\left(1-\frac{2b}{a} - e^{-2b/a}\right),$$ where ##k = \frac{1}{4\pi\varepsilon_0}##. Now expand the exponential term to second order, and you'll get the result you seek.
     
  6. Mar 21, 2017 #5
    Thanks vela. I see how that all fits together now.

    I'm still a little confused with this bit though:
    I know how to find the largest value of ##b## using a derivative (if that's what the questions is asking), but I'm not sure I understand the bit about 'one part in a thousand'. I'm thinking the idea behind this is to find the point where the Coulomb model first matches the ground state as ##b## increases from ##0##, which would be the maximum value ##b##?
     
  7. Mar 21, 2017 #6

    vela

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    The question is asking when do the unperturbed and perturbed energies differ by less than 0.1%.
     
  8. Mar 21, 2017 #7
    I think I know where I'm heading now, thanks.
     
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