# Is the given statement regarding Radial Probability Distribution Function correct or not?

Radial Distribution Function $$(4πr^2R^2(r))$$ gives the probability of the electron being present at a distance $$r$$ from the nucleus.

Answer: The given statement is correct.

My Query: According to me, isn't the Radial Probability Distribution Function ($$RPDF$$) used for finding the probability of an electron being present at a distance $$r$$ from the nucleus within a region of thickness $$dr$$?

Therefore, shouldn't the above statement be wrong?

• The emphasized part is implicitly assumed. Anyway, the probability to be exactly at a distance $r$ is $0$. Jul 1 '20 at 6:39
• You are correct it should be $r \to r+dr$ but it is not often written that way. Jul 1 '20 at 6:43
• @IvanNeretin are you saying that both answers are correct? Jul 1 '20 at 7:03
• @porphyrin so then mine is correct too, right? Jul 1 '20 at 7:03

Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance.

Technically, $$r^2R^2$$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $$r = r_1$$ and $$r = r_2$$ would therefore be

$$P = \int_{r_1}^{r_2}r^2R^2 \, \mathrm dr \tag{1}$$

For an infinitesimal region, it suffices to multiply by $$\mathrm dr$$, such that the infinitesimal probability of finding an electron in the region $$(r, r + \mathrm dr)$$ is

$$\mathrm dP = r^2R^2 \,\mathrm dr \tag{2}$$

Notice how if you integrate $$(2)$$ over a finite region you get $$(1)$$.

Also, some books give this extra factor of $$4\pi$$, which I never really understood. You have two choices: either the probability density function is $$r^2R^2$$ (which works for all orbitals), or $$4\pi r^2\psi^2$$ (which works only for spherically symmetric orbitals, i.e. s-orbitals). I've written about this in more detail at What is the exact definition of the radial distribution function?, so won't repeat myself here.

• I've read your answer before posting this question, and thanks a lot1 Jul 1 '20 at 7:02