Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Anyone know the answer to this? "the probability of finding electron density at the nucleus for a p orbital is"? A) 0.0 B)1.0 C) The same as an s-orbital D) 0.5

    Offline

    12
    ReputationRep:
    It's 0...but if you're doing stuff with orbitals you should have had that pointed out to you.
    Offline

    13
    ReputationRep:
    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: \psi = 2 (\frac{1}{4\pi})^\frac{1}{2} (\frac{Z}{a_0})^\frac{3}{2} e^-\frac{\rho}{2}

    2p: \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  \rho = \frac{2Zr}{a_0}

     a_0 is the Bohr radius, Z is the atomic number, r distance from nucleus.
    Offline

    14
    ReputationRep:
    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.
    Offline

    12
    ReputationRep:
    I was assuming that the question was basically asking what \psi \psi^* at the nucleus is for a p-orbital.
    Offline

    13
    ReputationRep:
    (Original post by 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 \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).
    Offline

    13
    ReputationRep:
    (Original post by Kyle_S-C)
    I was assuming that the question was basically asking what \psi \psi^* at the nucleus is for a p-orbital.
    Me too - either that or 4\pi r^2 |\psi \psi^*|
    Offline

    12
    ReputationRep:
    The question would suggest it just wants a probability distribution function rather than a rdf.
    Offline

    14
    ReputationRep:
    (Original post by =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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: August 31, 2007

University open days

  • University of Roehampton
    All departments Undergraduate
    Sat, 17 Nov '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 17 Nov '18
  • Bournemouth University
    Undergraduate Open Day Undergraduate
    Sat, 17 Nov '18
Poll
Black Friday: Yay or Nay?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.