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}
$
Best Answer
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.