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The modulus squared of the radial wave function gives the probability of finding an electron in an infinitesimal volume dv. On the other hand, the radial distribution gives the probability of finding an electron in a shell of radius r and thickness dr, or in other words, the probability of finding an electron at a radius r.

I know how the radial distribution is derived and it being 0 at the nucleus makes sense mathematically but not physically. The radial distribution being 0 at the nucleus means that an electron cannot be found in a shell of r=0 & thickness dr ( a sphere of radius dr), but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus. These results appear to contradict each other.

Is there something wrong in my physical interpretation of the 2 functions?

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You state 'but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus.'

To add to Poutnik's answer, I think that this sentence states the misunderstanding. The radial part of the wavefunction is $R_{n\ell}(r)$ and when normalised

$$\int_0^\infty R_{n\ell}^*(r)R_{n\ell}(R)r^2 dr=\frac{4}{a_0^3}\int_0^\infty e^{-2r/a_0}r^2 dr=1$$

where the $r^2$ comes from the volume element needed to integrate over all angles which is $r^2\sin(\theta)drd\theta d\phi$. The probability from $r\to r+dr$ is then for the 1s orbital $\displaystyle p=\frac{4}{a_0^3}r^2e^{-2r/a_0}dr$ whereas the 1s wavefunction itself has the form $\sim e^{-r/a_0}$. For other orbitals, e.g $2p_{\pm 1}$ the wavefunction is a function with complex numbers so cannot represent anything physical.

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For the wave function of $\mathrm{s}$ orbital $\psi(r)$, the radial probability is:

$P(r)=4\pi r^2 \cdot|\psi(r)|^2$.

So even if the differential probability density $|\psi(r)|^2$ is nonzero for $r=0$, the integral probability over a spherical surface $P(r)$ in $r,r+\mathrm{d}r$ interval raises with $r^2$ and is zero for $r=0$.

i understand the explanation mathematically but I still dont know what that means physically. Can an electron be found at the nucleus or not?

The current wave function based model says it can. But considering the relative size of the region and the WF value, probability is very low.

Also note 2 important implied considerations in the current quantum atomic model:

  • a point-like character of a nucleus
  • as the center of spherical coordinates is taken the atom mass center, instead of nucleus itself. So $r=0$ is not at a nucleus, but typically 100 times farther away from a nucleus, compared to a typical nucleus size.

In classical analogy, the surface of a zero-radius sphere contains no adsorbed compound, even if surface concentration in $\pu{g/m3}$ is nonzero.

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  • $\begingroup$ i understand the explanation mathematically but I still dont know what that means physically. Can an electron be found at the nucleus or not? $\endgroup$ – OVERWOOTCH Feb 2 at 9:55
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    $\begingroup$ @OVERWOOTH You are confusing the radial wave function with something that has physical meaning in reality, which is the radial probability distribution. That function says that the probability of finding the electron at within some infinitesimal $dr$ of the nucleus is zero. $\endgroup$ – Zhe Feb 2 at 13:28
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    $\begingroup$ @OVERWOOTCH ....the probability of finding the electron at within some infinitesimal dr of the nucleus is zero .......which is true for any other infinitesimal space element, it is not limited to the one at r=0.(this may not be obvious at first glance from these functions of spherical coordinates ) $\endgroup$ – Poutnik Feb 2 at 13:34
  • $\begingroup$ so it is not possible to find an electron within some infinitismal dr of the nucleus, but it is possible to find it in a cube of infinitismal volume dv at the nucleus (as expressed by the non zero wavefunction at the nucleus)?? Don't these 2 statements contradict eachother? $\endgroup$ – OVERWOOTCH Feb 2 at 13:58
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    $\begingroup$ @OVERWOOTCH No, as you misinterpret it.. "anything not infinite" times dV = 0, whatever coordinates is dV expressed with. Respectively, dP=psi^2.dV $\endgroup$ – Poutnik Feb 2 at 14:04

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