# What is the radial probability distribution function and what is its significance? [duplicate]

Below I have given three graphs for the $$\ce{1s}$$ orbital.

$$R(r)$$ is the radial part of the wave function of the electron. $$R^2(r)$$ is the radial part of the wave function of the electron multiplied by its complex conjugate which gives us the probability of finding the electron at some point. $$R^2(r)4πr^{2}dr$$ is what is called the 'radial probability distribution function'. It gives us the probability of finding the electron in a thin spherical shell of thickness $$dr$$.

Clearly from the second graph we can see that as $$r$$ increases $$R^2(r)$$ decreases which means that as we move away from the nucleus, the probability of finding the electron decreases. Now, let us look at the third graph. It is telling us that as we move away from the nucleus, the probability of finding the electron in a thin spherical shell of thickness $$dr$$ increases and it reaches a maximum value at some $$r$$ which we will call $$r_{max}$$. $$r_{max}$$ is the distance from the nucleus at which the probability of finding the electron is maximum. Now I have a couple of questions.

From the second graph, we know for sure that as we move away from the nucleus the probability of finding the electron decreases. But what the third graph is telling us is the probability of finding the electron increases and reaches a maximum value. But how is this possible? We just saw that the probability decreases with increase in distance from the nucleus. How can the probability attain a maximum value at some $$r$$ if it is decreasing with $$r$$? Aren't the two graphs contradictory to each other? Moreover it is a $$\ce{1s}$$ orbital and so by the electron density diagram we can tell for sure that the probability of finding the electron decreases with increase in distance from the nucleus. As it turns out, $$r_{max}$$ is equal to $$0.529×10^{-10} m$$. It means that at this distance from the nucleus probability of finding the electron is maximum. But we clearly know that as we move away from the nucleus the probability of finding the electron decreases. So how can the probability of finding the electron be maximum at $$r_{max}$$? The probability of finding the electron should be maximum just outside the nucleus and not at $$r_{max}$$, right?

• Be aware that $R^2(r)$ is probability density function, with probability $P_{r_1,r_2}=\int_{V}^{}{R^2(r) \cdot \mathrm{d}V}=\int_{r_1}^{r_2}{R^2(r) \cdot 4 \pi r^2 \cdot \mathrm{d}r}$ Therefore, even if probability density has maximum for r=0, probability for r,r+dr for r going to 0 converges to 0 as well. Commented Jun 8, 2021 at 15:03
• @Poutnik, but the probability of finding the electron should be maximum just outside the nucleus and not at $r_{max}$, right? Commented Jun 8, 2021 at 15:07
• What case has higher probability ? probability density=100 and volume = 1, or probability density 1 and volume = 1000 ? Maximal density for r=0 does not help if sphere surface converges to 0 faster than the desnity rises ? Commented Jun 8, 2021 at 15:09
• Calling two different graphs the same name ("probability") is a great way to get yourself confused, especially when you know for sure that they are different, and moreover, how exactly are they different. Commented Jun 8, 2021 at 15:11

Imagine a tango party with a large dance floor and a single porta-potty (one square meter floor area). There is a larger chance of finding people on the dance floor than in the restroom. On the other hand, there might be a larger chance of finding someone in the restroom than on a specific square meter area on the dance floor.

In other words, the two graphs address two different questions. The second shows you how likely it is to find an electron in a box of given volume closer or further from the nucleus. Perhaps surprisingly, the highest probability for the electron in a hydrogen atoms ground state is right at the nucleus. The third graph shows how likely it is to find an electron at a given distance. Because the volume available at a given distance increases with the square of the distance, you get a different shape of the curve. The third curve is relevant for calculating the mean or the average distance of the electron from the nucleus (or a surface within the electron is located inside with a probability of 90%).

Below are all three functions in the same graph (the y-axis is arbitrary). You can see that the radial part of the wavefunction (green) is different from its square (purple), which again is different from the radial probability density (orange). The x-axis should read $$r / a_0$$.

(source)

• Nice answer! But I have a question; (1)So, mathematically it does make sense why radial probability is maximum at a certain distance for (say) 1s orbital. But is there a physical explanation for this, shouldn't the electron be more stabilised just outside the nucleus? Commented Jun 9, 2021 at 2:03
• Yes @Rishi even I have the same question. Commented Jun 9, 2021 at 10:22
• @Rishi The potential is strongest near the nucleus, this is why the density is highest near the nucleus. The population density in New York State is highest in New York City (Manhattan, probably), but a majority of people live outside of the city. The average distance of people to New York Times Square places them outside of Manhattan as well. The conceptual hiccup comes with the math, not the physics or the chemistry.
– Karsten
Commented Jun 9, 2021 at 13:52
• Brilliant analogy @Karsten Theis! I have one last question : If someone asks where is the probability of finding the electron in a $\ce{1s}$ orbital maximum - at a distance of $0.529 × 10^{-10} m$ $(r_{max})$ from the nucleus or just outside the nucleus? Then what would we answer? It would mean the world to me if you could answer the above question. Commented Jun 14, 2021 at 16:12
• @user281837 It depends on the nuances in the question. If they ask "at what distance from the nucleus would I observe the electron, I would say at around $r_{max}$. If they are interested in scattering by electrons (like in crystallography), I would say the closer to the nucleus, the more scattering because the electron density gets higher as you approach the nucleus. Of course, that is specifically for the 1s, where we don't have any nodes..
– Karsten
Commented Jun 14, 2021 at 17:18