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Referring to the answer by DSVA (Most probable point for finding an electron in the 1s orbital of a Hydrogen atom)

There's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance)

I face a problem in solving the maximum probability of finding electron in a 2p orbital. $$\psi=R_{2,1}Y_{1,0}=\sqrt\frac{1}{32\pi a_o^3} \left(\frac{r}{a_o}\right) \exp\left(\frac{-r}{2a_o}\right) \cos\theta $$

Using probability density function, differentiate and equate to zero$$ \frac{d\psi^2}{dr}=constants\left(2r-\frac{r^2}{a_o}\right)\exp\left(\frac{-r}{a_o}\right)=0$$ I obtain$$r=2a_o$$ Using radial probability distribution, differentiate and equate to zero$$P(r)=r^2|R(r)|^2 $$ Referring Atkins' Physical Chemistry (pg. 312), it is stated that spherical harmonics is normalised to 1.$$\frac{dP}{dr}=constants\left(4r^3-\frac{r^4}{a_o}\right)\exp\left(\frac{-r}{a_o}\right)=0 $$ I obtain$$r=4a_o$$

My question: When should we use radial probability or probability density to find maximum probability of finding electron and its most probable distance? What does the difference of values mean?

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What you are interested in is the maximum probability of finding the electron at a given position.

In general, the probability of finding an electron in a volume $\pmb{V}$ (not necessarily infinitesimal) is

$$ \int_{\pmb{V}} |\Psi(\pmb{r})|^2 d^3 \pmb{r}$$

This integral is, so far, abstract, because we haven't defined a coordinate system yet. If you want to use spherical coordinates - which is a common thing to do -, then you can translate this integral to

$$ \int_\pmb{V} |\Psi(\pmb{r})|^2 r^2 \sin \theta dr d\theta d\phi$$

So if you want the maximum probability, then what you are interested is in maximizing the integrand with respect to $r$

$$ \max_r |\Psi(\pmb{r})|^2 r^2 \sin \theta$$

If you then take the derivative of this quantity with respect to $r$ and set it to zero, you end up with

$$ -\frac{r^3 e^{-\frac{2 r}{a_0}} \left(r-2 a_0\right) \sin (\theta ) \cos ^2(\theta )}{16 \pi a_0^6} =0$$

which has a solution indeed $r=2 a_0$.

A common mistake is finding the maximum of the wavefunction itself is the fact that it only takes into account the decay of the wavefunction with the distance from the nucleus, but it doesn't account for the volume of the infinitesimal disc at that range. Imagine if the wavefunction would be constant up to a certain critical radius $r_c$, and zero outside(I know it breaks quantum mechanics but don't think too hard about that for a second): the maximum probability of finding it would be right at $r_c$, because that radius has the largest geometrical probability to find the constant electron density in.

This is actually quite often misunderstood in the context of the definition of the Bohr radius: it is not the maximum of the electron density of the hydrogen atom, but of the probability density. I hope the distinction is clearer now.

(Two minor mistakes: I'm almost certain you need to include an angular factor in the probability density which you missed, and you shouldn't take the second derivative in the first case - which is irrelevant, anyways).

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  • $\begingroup$ I'm really sorry and would like to apologise if you have referred to my hydrogen-like wave function for 2pz orbital in the question. I made a mistake for the exponential term. It should be (-r/2a) instead of (-r/a), so I edited just now. Therefore, I take the derivative of the integrand you mentioned with the corrected wave function. I get (r-4a) instead. $\endgroup$
    – Student
    Jul 1 at 13:37
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    $\begingroup$ Yes, admittedly I just took your expression directly. This misprint will change the result probably, but not the explanation. $\endgroup$
    – Szgoger
    Jul 1 at 13:57

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