0
$\begingroup$

The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.

The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.

so if probability density cannot be zero, then what about nodes? Arent they regions where probabilty density is zero?

$\endgroup$
2
  • $\begingroup$ What is a "node" here? $\endgroup$
    – peterh
    Oct 6, 2020 at 13:46
  • $\begingroup$ A node is in the orbital context generally a 3D surface with the zero $|\Psi|^2$ value, what has interpretation as zero probability of an electron occurance. $\endgroup$
    – Poutnik
    Oct 6, 2020 at 15:04

1 Answer 1

2
$\begingroup$

The text is simplifying, as what it says is valid for and only for 1s orbital.

The text author had probably in mind that there is no sharp end of electron density as misinterpretation of orbital shapes may suggest. There is an asymptotic exponential-like ceasing of the wave function values toward infinity.

All higher s orbitals with n>1 have n-1 radial nodes, having zero $|\Psi|^2$ value of wave function at n-1 different radii.

Similarly, 2p orbitals have the node plane passing the center between the two lobes and upper p orbitals have additional node surfaces. Similar for d and f orbitals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.