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Kinetic energy of the hydrogen atom in its ground state

  1. Nov 12, 2016 #1
    I saw another post about this but i didn't quite find what i was looking for there so i thought i'd give it a go instead with a thread.

    1. The problem statement, all variables and given/known data
    Calculate the exact value of the kinetic energy of the hydrogen atom in its ground state. No more information is given, we are referred to litterature and stuff.

    2. Relevant equations
    I am thinking that my approach is wrong in the sense that i am trying to do something in a way that it's not allowed to. We have the hamiltonian [tex]\hat{H}\psi=E\psi[/tex] which in this case is [tex](-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{4\pi\epsilon_0r})\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)=-13.6eV\psi(r,\theta,\phi)[/tex] and the last term is because we know that it is in its groundstate.

    3. The attempt at a solution
    What i thought that i'd do that i now suspect is not "allowed" mathematically was to let the [itex]\psi(r,\theta,\phi)[/itex]s cancel themselves out and then since the first term on the left side represents the kinetic energy, just move the right term over to the right side and simply calculate the value. The value i get is incredibly small, since it is the hydrogenatom in its groundstate we can interpret the function with the values [tex]Z=1[/tex][tex]e=1.6*10^{-19}[/tex][tex]\epsilon_0=8.8*10^{-12}[/tex] and [tex]r=0.52Å=0.52*10^{-10}[/tex] and if i proceed to do as i intended the value for the kinetic energy i get is 13.6eV or so which is wrong, i've been nudge to believe that the kinetic energy should be [itex]\frac{1}{2}[/itex][tex]kinetic energy=-13.6+\frac{(1.6*10^{-19})^2}{(4\pi8.8*10^{-12}*0.52*10^{-10})}=-13.599999eV[/tex]

    Should my approach instead be to try to calculate [tex](-\frac{\hbar^2}{2\mu}\nabla^2)\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] given that [tex]\psi(r,\theta,\phi)_{100}=\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] How would i go about and do this?
     
  2. jcsd
  3. Nov 12, 2016 #2

    blue_leaf77

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    When it comes to quantum mechanics, one can only calculate the average of an observable to describe the theoretical result of a measurement on that observable. So, the word "exact" in the question is supposed to be understood as the "expression" of the average or expectation value of kinetic energy.
    By the way, although your approach is indeed wrong, I need to mention that you cannot add two quantities with different units. That's why you get that 13.599999.
     
  4. Nov 12, 2016 #3
    The question is indeed to calculate the exact value. This is the 2nd of 2 questions, where the first one is the calculate the value with the Thomas-Fermi approach and Hartree units, which gave me the value 0.3, and then the 2nd question, the one i am adressing above, is to instead calculate the exact value.
     
  5. Nov 12, 2016 #4

    PeroK

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    You know the total energy. Maybe if you could calculate the expected value of the potential energy, then you could get the expected value of the kinetic energy from that?
     
  6. Nov 12, 2016 #5
    Hmm, that approach might actually work! Of course i would do this like [tex]\langle\psi_{100}|potentialenergy|\psi_{100}\rangle[/tex] right? That's how i've calculated expected values before. I've just finished lunch, but i'll give that a try after and i'll get back to you guys, thanks for the help! :)
     
  7. Nov 12, 2016 #6

    PeroK

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    What you have here is essentially an operator equation:

    ##\hat{H}\psi = E \psi##

    or

    ##\hat{T}\psi + \hat{V} \psi = E\psi##

    What you can't do is cancel the ##\psi##. What you could do is take the inner product with ##\psi## to get an equation involving expected values:

    ##\langle \psi | \hat{T}\psi + \hat{V} \psi \rangle = \langle \psi | E\psi \rangle##

    ##\langle T \rangle + \langle V \rangle = E##
     
    Last edited: Nov 12, 2016
  8. Nov 12, 2016 #7
    I'll get to that right now :) Get back to you in a second
     
  9. Nov 12, 2016 #8
    I still don't get it quite right it looks like. [tex]\langle\psi|V(r)|\psi\rangle=\langle\psi|-\frac{k}{r}|\psi\rangle=\frac{-k4\pi}{\pi a_0}\int_{0}^{\infty} re^{\frac{2r}{a_0}} dr[/tex] which i get to become [tex]-ka_0^2=-\frac{-Ze^2a_0}{4\pi\epsilon_0}=\frac{-(1.6*10^{-10})^20.52*10^{-10}}{4\pi8.8*10^{-12}}=-1.2*10^{-38}[/tex] I must have done something wrong somewhere. The 4pi early in the numerator came from the [itex]sin(\theta)d\theta d\phi[/itex] in the spherical coordinates integral.
     
  10. Nov 12, 2016 #9

    PeroK

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    What wave function are you using? Also, I would break this problem down and first calculate:

    ##\langle \frac{1}{r} \rangle##

    Whay do you get for that?
     
  11. Nov 12, 2016 #10
    The wavefunction i am using is the one for hydrigen in its m=0, l=0 and n=1-state, that is [itex]\frac{1}{\sqrt{\pi a_0}}e^{\frac{-r}{a_0}}[/itex]
     
  12. Nov 12, 2016 #11

    PeroK

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    It should be:

    [itex]\frac{1}{\sqrt{\pi a_0^3}}e^{\frac{-r}{a_0}}[/itex]

    By the way, you may not need any numbers for this problem at all.
     
  13. Nov 12, 2016 #12
    Now i see it should ba 3 there.... this vastly changes things.. ! I'll do the math once again for the expected potential and i'll get back!
     
  14. Nov 12, 2016 #13
    For this i get [tex]\frac{1}{a_0}=1.92*10^{10}[/tex] which if i later also add in the rest of the equation for the potential energy gives me that the expected value for the potential energy should be [itex]-4.45*10^{-18}[/itex] which is wrong i guess.. :/
     
  15. Nov 12, 2016 #14

    PeroK

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    No numbers allowed! How is ##a_0## defined?
     
  16. Nov 12, 2016 #15
    [tex]a_0[/tex] is the bohr-radius
     
  17. Nov 12, 2016 #16

    PeroK

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    Yes, okay, but how is it related to the potential ##V(r)##?
     
  18. Nov 12, 2016 #17
    The way i'd like to see it be is that the potential at the inner most radious for the electron (ground state) should be 0 because it can't jump back to a lower state but i guess this is not the case.. :/
     
  19. Nov 12, 2016 #18

    PeroK

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    ##a_0 = \frac{4\pi \epsilon_0 \hbar^2}{me^2}##

    You could have looked that up online.

    ##V(r)## can be expressed, therefore, in terms of ##a_0##

    Also, you'll need ##E_0## in terms of ##a_0##.
     
  20. Nov 12, 2016 #19
    Oh i didn't understand your question earlier, let me dig into this for a while, i'll get back to you
     
  21. Nov 12, 2016 #20

    vela

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    That's actually the right answer, but that's not the answer you got. You got -13.6 eV, which is wrong because the kinetic energy shouldn't be negative. You can't just erase the minus sign because you don't want it there.
     
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