# The probability of finding the electron in the H-atom

In the book Arthur Beiser - Concepts of modern physics [page 213] author separates the variables in the polar Schrödinger equation assuming:

$$\psi_{nlm}=R(r)\Phi(\phi)\Theta(\theta)$$

then there a statement that the differential od space in the polar coordinate system is:

$$dV=(dr)\cdot (d\theta r)\cdot (r\sin\theta d\phi)$$

I understand this, but on the next page there is a statement:

As $\Phi$ and $\Theta$ are normalised functions, the actual probability $P(r)dr$ of finding the electron in a hydrogen atom somewhere in the spherical shell between $r$ and $r+dr$ from the nucleus is:

$$P(r)dr=r^2|R(r)|^2dr\,\int\limits_{0}^{\pi}|\Theta(\theta)|^2\sin\theta d\theta \, \int\limits_{0}^{2\pi}|\Phi|^2 d\phi=r^2|R(r)|^2dr$$

In this equation i can recognize the differential of volume described above and the wavefunction $\psi_{nlm}=R(r)\Phi(\phi)\Theta(\theta)$. I also know that normalization of the angular functions over the angles returns 1, but I don't understand why there is no integration of the radial part... Can anyone explain a bit?

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There is no integration of the radial part because, as you said yourself, we want the probability of finding the electron somewhere in the spherical shell between $r$ and $r+dr$ from the nucleus. (in a differential shell between $r$ and $r+dr$, and no need to integrate over $r$.)

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Thank you very much. So if we are satisfied with a fraction of a probability there is no need to integrate :) What confused me was that feeling that if we want to integrate the equation we have to do it on both sides and not only one side... It is weird to me that we integrate just part of the equation. – 71GA Aug 15 '13 at 14:37
I have one more question about this. Why did we use the diferential of volume when we are searching for $P(r)dr$ – 71GA Aug 15 '13 at 14:41
Never mind i think i understand :) – 71GA Aug 15 '13 at 14:46
If you are looking for $P(r)dr$ you have to find $P(r)$ first (from $P(r,\theta,\phi)$) . – Mostafa Aug 15 '13 at 15:00
How do i do this? – 71GA Aug 15 '13 at 20:03

$P(r)dr$ gives you only the probability in an infinitesimal spherical shell around the center. The integration you're expecting is made, when you want to know the probability in a non-infinitesimal shell around the center.

For example, you'd like to know what is the probability of finding an electron between $r=1$ and $r=2$ (in whatever coordinates), you'd integrate $$P(2<r<1) = \int_1^2 P(r)dr$$.

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