The Student Room Group
Reply 1
It's 0...but if you're doing stuff with orbitals you should have had that pointed out to you.
Reply 2
Hmm, a tricky one. The probability density (value of the RDF) must obviously be zero as there are no electrons in the nucleus. This is true for all orbitals, so in a way you could argue both A) and C) are correct.
The difference is, of course, that the value of the 2p wavefunction is also zero at the nucleus whereas it is actually maximum for s-orbitals!

Out of interest, the wavefunctions for 1s and 2p in a hydrogenic atom are as follows:

1s: ψ=2(14π)12(Za0)32eρ2\psi = 2 (\frac{1}{4\pi})^\frac{1}{2} (\frac{Z}{a_0})^\frac{3}{2} e^-\frac{\rho}{2}

2p: ψ=(14(6)12)(516π)12(3cos2θ1)(Za0)32ρeρ4\psi = (\frac{1}{4(6)^\frac{1}{2}}) (\frac{5}{16\pi})^\frac{1}{2} (3\cos^2 \theta - 1) (\frac{Z}{a_0})^\frac{3}{2} \rho e^-\frac{\rho}{4}

where ρ=2Zra0 \rho = \frac{2Zr}{a_0}

a0 a_0 is the Bohr radius, Z is the atomic number, r distance from nucleus.
The way I understand the electron density of the s-orbital is that there is a probability of finding an electron in the nucleus, but we have to remember what the electron density actually is, the probability per point in space. When we consider the relative number of points in space (i.e. the volume) occupied by the nucleus compared to the total volume occupied by the electron we realise the although the probability of finding the electron is high at the nucleus compared to a point is space outside, there are trillions more points in space outside the nucleus than inside, hence why electrons don't wind up in the nucleus that often. Sorry that's a bit waffly.

Electrons do go into the nucleus and under specific circumstances can be captured by a proton to form a neutron and emit a neutrino and reduce the atomic number by 1, so called "K-shell capture" a form of radioactive decay.
Reply 4
I was assuming that the question was basically asking what ψψ\psi \psi^* at the nucleus is for a p-orbital.
Reply 5
ChemistBoy
The way I understand the electron density of the s-orbital is that there is a probability of finding an electron in the nucleus, but we have to remember what the electron density actually is, the probability per point in space. When we consider the relative number of points in space (i.e. the volume) occupied by the nucleus compared to the total volume occupied by the electron we realise the although the probability of finding the electron is high at the nucleus compared to a point is space outside, there are trillions more points in space outside the nucleus than inside, hence why electrons don't wind up in the nucleus that often. Sorry that's a bit waffly.
Electrons do go into the nucleus and under specific circumstances can be captured by a proton to form a neutron and emit a neutrino and reduce the atomic number by 1, so called "K-shell capture" a form of radioactive decay.


Sure, almost anything can happen under specific circumstances but if you consider a stable isotope in its ground state then the absolute probability of finding an electron right in its nucleus is zero according to contemporary quantum mechanics. In a p-orbital there is actually a nodal plane going through the nucleus! That is, if you assume that the nucleus is infinitely small, there will be SOME very small probability if you consider a certain thickness dτ\mathrm{d}\tau.
I'm slightly confused about s-orbitals though, the RDF is zero at r=0 but the wavefunction is not (unlike for all other orbitals).
Reply 6
Kyle_S-C
I was assuming that the question was basically asking what ψψ\psi \psi^* at the nucleus is for a p-orbital.


Me too - either that or 4πr2ψψ4\pi r^2 |\psi \psi^*|
Reply 7
The question would suggest it just wants a probability distribution function rather than a rdf.
=gabriel=

I'm slightly confused about s-orbitals though, the RDF is zero at r=0 but the wavefunction is not (unlike for all other orbitals).


Probably because you assume that the nucleus is a infinitely small point when calculating the radial distribution function.

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