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When we want to talk about the most probable radius to find the electron in $1\text{s}$ orbital, why do we talk about the radial density and not the probability itself? For instance, the probability of finding the the electron at a radial distance $r$ is ($\rho$ being the square modulus of the radial part of wavefunction):$$ \mathrm{d}P~=~\rho\left(r\right) \, \mathrm{d}V ~=~ \rho\left(r\right)4\pi r^2 \, \mathrm{d}r ~=~\rho' \left(r\right) \, \mathrm{d}r \,,$$ since the electron can be anywhere in the shell at radius $r$. Now, the probability of finding electron is maximum when $\frac{\mathrm{d}P}{\mathrm{d}r}=0$, right? Isn't that the condition for maxima? But that gives $r=0$, i.e. the nucleus. Now I know that this is wrong, I'm supposed to get the Bohr radius.

However, and we all know this, extremising $\rho' \left(r\right)$ i.e. the radial density gives the correct answer.

What I don't understand is, why am I wrong the first way? After all, maximising (okay, extremising) anything is to put the first derivative to zero, right? Then what am I missing here? I can 'see' that this is wrong because $\frac{\mathrm{d}P}{\mathrm{d}r}=0$ means my radial density is zero, which is a little absurd, but that is not a good argument at all.

In a nutshell, why do we extremise the RADIAL DENSITY, and NOT PROBABILITY while finding the maximum probability? If we were to tell the most probable radius where this DENSITY is maximum, then I'd be fine with the second way. But if we were to find the greatest probability itself, isn't the second approach counter intuitive?

I know I am missing something very subtle; please point it out to me.

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The probability to find the electron in a volume is the product of that volume with the probability density. In your example the volume is a thin shell determined by $r-dr \lt r \lt r+dr$ with small $dr$. The probability to find the electron there is $P(r) = 12 \pi r^2 dr \cdot \rho(r) $. For a 1s orbital $\rho ( r ) $ is maximal at $r=0$, but for small $r$ the volume grows faster than $\rho ( r ) $ falls off. Therefore $P(r)$ is maximal at non-zero $r$.

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The problem you're having is that $P$ isn't what you think it is. The biggest clue is that $dP/dr$ is always positive for $r > 0$, whereas the actual probability should go to zero as $r$ approaches infinity, so the derivative should be negative at large $r$.

If we integrate the equation $$dP = \rho'(r)dr$$ we should choose limits for the integration. For example $$P(a) = \int dP = \int_0^a \rho'(r)dr = \textrm{probability}(r<a).$$ So, $P(r)$ is the cumulative probability, that is, the probability of finding the electron at a distance less than $r$ from the nucleus.

The actual probability density at $r$ is given by the derivative of the cumulative function: $$\textrm{probability density}(r) = \frac{dP}{dr} = \rho'(r).$$ This means that you want to maximize $\rho'(r)$ to find where the electron spends most of its time.

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You are confused between radial probablity function and probability density. The probability density would give you the highest value at $r=0$ but the probability function gives you the Bohr radius as its maxmium (for hydrogen atom). And when you compute the probability density (or in other words the electron distribution density) you are calculating the likelihood of finding electron in the vicinity of a PARTICULAR POINT, which btw hurts the wave-particle duality characteristic of an electron. The results is true buttt it is inconsistent with a shell model. And the result, in my opinion, is trivial because we all know that the electron will most likely to be found near the nucleus and less likely as $r$ increases. Dont you agree? We turn ourself to the radial distribution function to get free from the limitation of a volume element. A shell! When you integral the radial probability density over a range of r, you are actually computing the probability of finding electron in a shell of a certain thickness. That would greatly improve your chance of finding an electron!

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