# Why do we need the radial probability distribution function?

I know that the radial probability distribution function gives us the probability of finding an electron between in a spherical shell of thickness $$dr$$ which is at a distance $$r$$ from the nucleus. Now my question is : How is this function useful? For e.g., if we see the $$\Psi^2(r)$$ v.s. $$r$$ graph of $$1s$$ orbital of hydrogen atom, we can clearly see that the probability of finding the electron is maximum just outside the nucleus. But when we look at the radial probability distribution function it tells us that the probability of finding an electron is maximum at a distance of $$0.529 × 10^{-10}$$ m which is also known as Bohr's radius. This is because radial probability distribution function measures the probability of finding the electron in a spherical shell. Now obviously as the surface area of the spherical shell increases the probability of finding the electron will increase as we will be finding the electron in a larger volume. But the problem is that the probability of finding the electron is still maximum just outside the nucleus but yet the radial probability distribution function is maximum at Bohr's radius just because the volume is larger there. How does this help us? If we already know that the probability is maximum outside the nucleus by $$\Psi^2(r)$$ v.s. $$r$$ graph what was the need for this new function? Moreover this function accounts for the surface area and hence even if the probability of finding the electron is low somewhere and the surface area is high then also the graph will peak. Thus the graph does not even give us whether the probability of finding the electron is maximum or not as it also accounts for the surface area of the spherical shell. If it does not give us the correct information about the surface area then why do we need this function?

Also, as this function measures the probability of finding an electron between in a spherical shell of thickness $$dr$$ which is at a distance $$r$$ from the nucleus, doesn't this mean that the function will be useful only for s-orbitals as they are the only spherically symmetric orbitals? The other orbitals are not spherical and have lobes. How will this function help in those orbitals?

• I'm a little confused by your question(s), for the other non spherical orbitals (s, p, d etc subshells), they correspond to the solutions of the angular components. For the other radial solutions they correspond to other shells (corresponding to quantum number n). The other orbitals you talk about are composed of a Radial function and the other angular (orbital, azimuthal) functions. is there something I am missing that you are asking? Aug 12, 2021 at 4:44
• The following link gives the standard explanation on how the separable solutions to Schrodinger equation gives the n,l,m,s quantum numbers. chem.libretexts.org/Bookshelves/… is there something in it that you would consider missing? Aug 12, 2021 at 4:48
• Here's an idea: if you try to write a back of the envelope calculation that involves the average "size" of an atom, you wouldn't care that the point with maximum probability in space is at the center of the atom, just like in thermodynamics you don't care that the most probable state to find a system is the ground state because of $\exp(-E/KT)$: there are a lot more states with higher energy which contribute more to the system's average behavior. Aug 12, 2021 at 5:01
– Gert
Aug 12, 2021 at 6:35
• This is a rather silly, verbose question. $\mathrm{H}$ takes on the wave functions $\psi=R(r)Y(\varphi, \theta)$. Taking the modulus squared gives Born's $|\psi|^2$. Obviously $|R(r)|^2$ has its uses.
– Gert
Aug 12, 2021 at 7:01

For e.g., if we see the $$|\Psi(r)|^2$$ graph of 1s orbital of the hydrogen atom, we can clearly see that the probability of finding the electron is maximum just outside the nucleus.

It's not enough to say, just outside the nucleus. It's like saying a function that starts out from origin and goes down to zero at infinite has a maximum in between.

The below is the plot of $$|R(r)|^2$$ for $$n=2$$. It's clear that it's non-trivial. We found that there is a number of maxima. We also see the relative probability. Above all, it's a complete distribution. It tell far more than just finding the electron is maximum just outside the nucleus.

It's often useful that one wants to look at $$\langle r^2\rangle$$, once you know the radial probability distribution these things are trivial to find. For instance, In Larmor diamagnetism, if you imagine that $$\mathbf{B}$$ is applied in the z-direction, the expectation of the diamagnetic term of the Hamiltonian can be written as
$$\delta E=\frac{e^2}{8m}\langle |\mathbf{B}\times \mathbf{r}|^2\rangle =\frac{e^2B^2}{8m}\langle x^2+y^2\rangle$$ Using spherical symmetry $$\langle x^2+y^2\rangle =\frac{2}{3}\langle r^2\rangle$$

Now obviously as the surface area of the spherical shell increases the probability of finding the electron will increase as we will be finding the electron in a larger volume.

This is wrong! $$|R(r)|^2dr$$ gives the probability of finding electron in between $$r$$ to $$r+dr$$, there is no reason why this should increase and it doesn't as you see, it's necessary that $$|R(r)|^2\rightarrow 0$$ as $$r\rightarrow \infty$$.