3
$\begingroup$

I was reading notes from my first class in Quantum Physics that I received and left confused at the following statement:

For each principal quantum number $n$, the orbital set with the highest $\ell$ (orbital angular momentum quantum number) has its maximum electron density at the corresponding Bohr radius $n^2a_{0}$.

I just couldn't understand how the highest $\ell$ number corresponded to the Bohr radius i.e. maximum electron density of the $2$p orbital was calculated to be at $4a_{0}$.

$\endgroup$
2
  • $\begingroup$ Are you looking for an explanation other than just look at the wavefunctions and calculate where maximum is? $\endgroup$
    – AHusain
    Dec 28, 2018 at 22:12
  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Dec 28, 2018 at 22:55

1 Answer 1

1
$\begingroup$

Short of actually doing the calculation here is a crude estimate (which happens to be exact):

  1. Orbitals with maximal angular momentum $\ell=n-1$ (for given $n$) is expected to have a relatively well-defined notion of radius, cf. e.g. my Phys.SE answer here.

  2. Due to the virial theorem we have $\langle \frac{1}{r} \rangle=\frac{1}{n^2a_0}.$ It is then natural to expect OP's sought-for equation $\langle r \rangle=n^2 a_0.$

  3. For a more refined argument, see e.g. my Phys.SE answer here.

$\endgroup$

Your Answer

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

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