# 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?

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 $r$.)
• 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
• 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
$P(r)dr$ 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 $$P(2<r<1) = \int_1^2 P(r)dr$$.