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July 26, 2014

July 26, 2014

Posted by **Jill** on Tuesday, April 28, 2009 at 9:00pm.

- Quantum Mechanics -
**Count Iblis**, Tuesday, April 28, 2009 at 10:02pmThe 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 -
**Count Iblis**, Tuesday, April 28, 2009 at 10:04pmTypo:

Then the integral is

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

4/3 (R/a0)^3

- Quantum Mechanics -
**megan**, Tuesday, April 28, 2009 at 11:10pmi 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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