How is the wavefunction of an electron related to the atomic orbital? Why is it that wavefunction $\psi$ is maximum at the nucleus for $1 \text{s}$ orbital, even if the probability of finding electron is zero there. What is the significance of wavefunction. Can anyone explain it to me in simpler terms?
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\begin{array}{cccc}
& \textbf{Figure 1}.~~\psi~\text{vs.}~r \text{.}
& & \textbf{Figure 2}.~~4 \pi {r}^{2} {\psi}^{2}~\text{vs.}~r \text{.} \\[-100px]
\hspace{25px} & \hspace{250px} & \hspace{25px} & \hspace{250px}
\end{array}
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 A: The double-slit experiment led us to think that the intensity of a classical wave is proportional to the probability of finding the particle there.
Consequently, the wavefunction is such that its squared modulus represents the probability
$||\psi(x_0) || ^2 dx $ represents the probability of finding the particle at $x=x_0$ (in an infinitesimal environment of $x_0$)
The integral of that quantity
$$ \int_a^b  ||\psi(x) || ^2 dx $$
represents the probability of finding the particle in the interval $a,b$. It is usually written as
$$ \int_a^b  \psi^*(x)\cdot\psi(x)\cdot dx $$
which is the same.
The integral along the whole space means, obviously, the probability of finding it anywhere, so it must be 1, that's why $\psi$ mus be "normalized".

Finally, the $1s$ orbital is just the solution of the equation. It turns out to be like that. You have to solve an ideal hydrogen atom, that is, a Coulomb's potential. You find out that the radial functions of hydrogen are like that.
However, probability being maxium does not mean that the electron is always there.
A: For electrons in the $1s$ orbital, the wavefunction does, in fact, prescribe that the maximum probability of the position of the electron is at $r=0$. But that doesn't say that the electron is necessarily inside the nucleus: the integral of the position probability over an infinite range is 1, but the probability for $r=0$ is definitely not $1$.
For your second chart, you need to be more careful: $4\pi r^2$ is the area of a sphere of radius $r$, so $4\pi r^2|\psi|^2$ is the product of the area of the sphere and the probability of the electron lying on that sphere. Clearly for $r=0$, it should show zero. It's called a radial distribution function, and it's useful when you're trying to consider the variation of the electron being at a certain location (probability density) for a fixed distance from the nucleus, but not the probability of the electron being at a certain distance from the nucleus. The maximum is where it is because that's where you get an optimization of area of the sphere and probability of the radius.
Electron capture is a pretty cool phenomena: an electron (usually from $1s$ or $2s$) interacts with a proton from the nucleus to yield a neutron and an electron neutrino (note the conservation of charge). It's observed only in atoms with larger nuclei (I think certain isotopes of aluminium and holmium show it), and results in a change in atomic number, because you end up with one proton less, effectively!
This page has excellent charts which describe the probabilities of different positions for different orbitals.
