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 =
Then the integral is
4/3 pi R^3 |psi(0)|^2 =
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??