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I have a question about a task I am supposed to solve :

A hydrogen atom is in the ground state, with the radial part of the wave function equal to $R (r) _{n,l}=2 \space r_{B}^{-3/2}e^{\frac {-r}{r_{B}}}$. At which distance (from the nucleus) is the radial-probability density highest?

Here the H atom wave function is $\psi (r,\theta,\phi)=Y_{ml}(\theta,\phi)\, R (r) _{n,l}$ and $r_B =0.0528 \:\rm nm$.

My understanding:

The radial-probability density equals $(R (r) _{n,l})^2=4 \, r_{B}^{-3}e^{\frac {-2 \space r}{r_{b}}}$ and, because $r\in [0,\infty]$, the maximum should be at $r=0$? The textbook says it's at $r_B$.

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  • $\begingroup$ Hint: The probability density can be defined as the function $f(r)$ such that the probability of finding the electron between $r$ and $r+\mathrm{d}r$ is $P(r) = f(r)\mathrm{d}r$. Think hard about this, and compare to what you have done or not done. $\endgroup$ – garyp Aug 26 '16 at 12:51
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The radial probability density distribution is not given by your formula but by:

$$4\pi r^2R^2(r)_{n,l}$$

Source and explanation.

So the probability density at $r=0$ is in fact zero.

See for example for the $1s$ orbital.

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  • $\begingroup$ of course... how could I miss that... 4πr2 comes from the "sphere part".... Thank you very much! $\endgroup$ – Luka8281 Aug 26 '16 at 13:27
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This is a bit tricky at first, but essentially the difference is that $|\psi(r,\theta,\phi)|^2$ gives you the probability density at a unit volume at the given point, while your book is calculating the likeliest radius by itself, so you need to include the fact that a spherical shell of thickness $\mathrm dr$ has a bigger volume at bigger $r$.

Thus, you need to maximize the probability density over observations of the radius, which includes the volume element of that sphere, $$ 4\pi r^2 |R_{nl}(r)|^2 = \frac{16\pi}{r_B^3} r^2 e^{-2r/r_B}, $$ and that has a maximum at nonzero $r_B$. More intuitively, even if per unit volume the electron is most likely to be found near the nucleus, there's just not a lot of volume that's that close to the nucleus.

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  • $\begingroup$ It's basically $\int_{0}^{\pi}d\phi \space \int_{0}^{2\pi}|\psi|^{2} r^{2}\sin \theta \space d\phi \space d \theta$, So it tells me at which $r$ I am most likely to detect the electron, regardless of $\theta $ and $\phi$, right? $\endgroup$ – Luka8281 Aug 26 '16 at 13:37
  • $\begingroup$ The $4\pi$ part should actually vanish, since $Y(\phi, \theta)$ are normalised? $\endgroup$ – Luka8281 Aug 26 '16 at 13:46
  • $\begingroup$ @Luka The constants don't particularly matter, because you're just looking for the maximum and that doesn't change with a re-scaling. You're right, though, that what you're maximizing is the integral you pose, and that the $4\pi$ vanishes with the $Y_{lm}$ normalization. I left it in to make it clearer what the origin was, and because it's harmless for the calculation you were doing. $\endgroup$ – Emilio Pisanty Aug 26 '16 at 13:59

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