Can we view the infinite deep well, as a infinite barrier?
only from the point of view of r=0, but the electron is not there in the classical model; and even in the Schrodinger model r=0 is only one infinitesimal point, you would have to considered the expectation value of the energy.
what would the wavefunction intersect , does it pass through (0,0)?
the s-orbitals do have a non-zero probability density at (0,0), but again that is only one point.
I'm quite confused by the idea of having negative energy, does it mean I need infinite energy to get out of the well?
No, the electron does not need infinite energy to get out of the well because it does not start at the bottom of the well in the classical model, and has zero probability of starting at the bottom of the well in the Schrodinger because it is only a single point.
You need to consider potential energy, kinetic energy and total energy. If you look at the following question and answer you should have a solution to the problem: Hydrogen atom: potential well and orbit radii
To find this Fourier transform, you start with
$$\Psi(\vec{p})=\frac{1}{(2\pi \hbar)^{3/2}}\int{e^{-i\vec{p}\cdot\vec{r} / \hbar }\psi(\vec{r})\mathrm d\vec r},$$
as you've written, but you need to be a bit more careful after that.
To begin with, you can't assume that $\vec p$ is along the $z$ axis, because $\vec p$ has been given to you as the argument of $\Psi(\vec p)$, and you can't change it. What you can do, however, is set up the coordinate system for the $\vec r$ integration so that its axes simplify your life, i.e. with $z$ along the direction of $\vec p$.
(The reason you can do this, by the way, is because the ground state $\psi(\vec r)$ is spherically symmetric. If it's not symmetric, such as e.g. anything that's not an $s$ state, then you need to use a decomposition of the plane wave into partial (spherical) waves, like the one in Jackson, 2nd ed., p. 471, eq. (10.43).)
Once you've done that, then yes, you can write $\vec p = (0,0,p)$, and you can further use spherical polar coordinates $\vec r = r(\sin(\theta) \cos(\phi), \sin(\theta)\sin(\phi), \cos(\theta))$ for your position vector, write
\begin{align}
\Psi(\vec{p})
& =
\frac{1}{(2\pi \hbar)^{3/2}}
\int_0^\infty \int_0^\pi \int_0^{2\pi}
e^{-ipr\cos(\theta) / \hbar }\psi(\vec{r})
\:r^2\sin(\theta)\:\mathrm d\phi\mathrm d\theta\mathrm dr
\\& =
\frac{1}{(2\pi \hbar)^{3/2}}\sqrt{\frac{1}{\pi a_{0}^{3}}}
\int_0^\infty \int_0^\pi \int_0^{2\pi}
e^{-ipr\cos(\theta) / \hbar }e^{-r/a_0}
\:r^2\sin(\theta)\:\mathrm d\phi\mathrm d\theta\mathrm dr,
\end{align}
and integrate away.
Alternatively (and this is really for any future visitors who chance upon here looking for an actual answer for the title question), you can cheat and look up the answer, which is found in e.g.
The Momentum Distribution in Hydrogen-Like Atoms. B. Podolsky and L. Pauling. Phys. Rev., 34, 109 (1929).
(Not you, Marc, by the way. Go and do the integral.)
Best Answer
Who says that it is impossible for the electron in its $2s$ state to be at $r=0$ ? The function $R_{2s}(r)$ you have correctly written reaches its maximum at $r=0$ !
As for $r=2a_0$ : the electron in an $s$ state, which means that the angular component $Y(\theta, \phi)$ is just one, no angular dependence at all.
The wave function for an electron in states $1s$ is positive everywhere, because again there is no angular dependence
$$ \psi_{1s}(r, \theta, \phi)\equiv R_{1s}(r) = 2 \frac{1}{\sqrt{a_0^{3}}} e^{-\frac{r}{a_0}}$$
But the $1s$ and $2s$ states are eigenfunctions of the Hamiltonian for different values of the energy. They must be orthogonal. Therefore, since there is no angular dependance at all, only the integration over $r$ can lead to a zero value. This means that $R_{2s}(r)$ cannot keep a constant sign, it must change sign at least once. In fact it turns out that it changes sign only once. (For any principal number $n>1$ the same argument shows $R_{ns}(r)$ must change sign at least once, but in that case it happens $(n-1)$ times, more than once for all $n \ge 3$).
Detailed calculations shows it happens for $r=2a_0$, but the exact value is irrelevant for your question : the fact that $R_{2s}(r)$ must change sign somewhere, and therefore assume a zero value for some value of $r$, is an inevitable consequence of its orthogonality with $R_{1s}(r)$ which is positive everywhere.