Quantum Mechanics

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Calculate the probability of finding an electron in a 1s H-atom orbital inside the nucleus. Show your work and state your assumptions.

  • Quantum Mechanics -

    The wavefunction of the hydrogen atom is:

    psi(r) = 1/sqrt(pi) a0^(-3/2) exp(-r/a0)

    To find the probability of the electron in the nucleus, you need to compute the integral:

    Integral from r = 0 to R of

    dr 4 pi r^2 |psi(r)|^2

    where R is the radius of the nucleus. Since R << a0, you can replace psi(r) by the value it assumes at r = 0. Then the integral is 4/3 pi R^2 |psi(0)|^2 =

    4/3 (R/a0)^3

  • Quantum Mechanics -

    Typo:

    Then the integral is

    4/3 pi R^3 |psi(0)|^2 =

    4/3 (R/a0)^3

  • Quantum Mechanics -

    i think you must be in my pchem class..

    i solved this problem differently from above:
    I used the same method as example 7-2 and homework problem 7-9 in McQuarrie. I assumed that since the nucleus consists of just one proton, the radius of the nucleus is the radius of a proton which is on the order of 1x10^-15 m.
    Ao represents the Bohr radius which is 5.29177x10^-11, so i used that to solve for the radius in terms of Ao:
    Rnuc = 1.89x10^-5*Ao
    Then i just used the exact same method from those problems to solve for the probability.
    The probability I got was zero, but there may be some very small probability beyond the sensitivity of my calculator?
    It kind of makes since though, i guess, because, since the nucleus is a proton, the electron could never really be inside the nucleus. ??
    thoughts?? or maybe i made a mistake??

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