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Average momentum squared of Psi(100) of hydrogen atom

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Calculate <p2> for ψ100 of the hydrogen atom

    2. Relevant equations

    ψ100 = 1/(√pi) (1/a0)3/2 e-r/a0

    0 r n e-B rdr = n!/Bn+1

    p2 = -hbar ∇2 = -hbar2 (r2 d2/dr2 +2 r d/dr) (ψ does not depend on ø or θ)

    3. The attempt at a solution


    <p2> = ∫ψ*(p2ψ)dV
    ∫dV = 4pi0r2dr

    <p2> = ∫1/(√pi) (1/a0)3/2 e-r/a0 (-hbar2 (r2 d2/dr2 +2 r d/dr) 1/(√pi) (1/a0)3/2 e-r/a0)dV

    <p2> = -4 hbar2/ a03 ∫r2 e-r/a0 ( (r/a0)2 - 2r/a0 ) e-r/a0 dr

    <p2> = -4 hbar2/ a03 ∫e-2r/a0 ( r4/a02 - 2r3/a0 ) dr


    The problem that I am running into is that I am calculating the integral to be 0:

    B= 2/a0 n1= 4 n2=3

    ∫e-2r/a0 ( r4/a02 - 2r3/a0 ) dr = (4!/ (a02 (2/a0)5) - (2*3!/ (a0 (2/a0)4 = (24a03/32 - 12a03/16) =0

    Am I doing something wrong, because I don't think <p2> would be 0 (that would indicate that the momentum is always 0)?
     
    Last edited: Apr 17, 2014
  2. jcsd
  3. Apr 17, 2014 #2

    Simon Bridge

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  4. Apr 17, 2014 #3
    Thank you. I left out the 1/r^2 in the laplace operator. When I redid the problem after fixing the operator, the answer I am now getting makes sense.
     
  5. Apr 17, 2014 #4

    Simon Bridge

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    Easy to do - there are so many examples where you don't need it.
     
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