Where would you not find an electron in a hydrogen atom?

Question

I am given the radial component of the wavefunction for an electron in its $2s$ state:

$$ R(r) = \frac{1}{2 \sqrt{2 a_0^{3}}} \left[ 2 - \frac{r}{a_0} \right] e^{-\frac{r}{2a_0}}$$

Such that its complete wavefunction is given by $\psi(r, \theta, \phi) = R(r) \times Y(\theta, \phi)$ where $Y(\theta, \phi)$ is the angular component and has no $r$ dependence.

Apparently, it is impossible for the electron to exist as at either $r=0$, $r=\infty$, or $r=2a_0$. I understand why it could not exist at $r=\infty$ as the exponential would tend to $0$, bringing the entire wavefunction to zero and the probability function with it. But I can't understand why $r=0$ or $r=2a_0$ would be a forbidden by this equation. My only guess for the second one would be that it has something to do with bringing that exponential down to $e^{-1}$.

Answer 1

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.

Answer 2

The probability relates to the wavefunction times the volume element which is proportional to $r^2dr$, giving you $r^2R_{nl}(r)$.

So at $r=0$ the value you want is zero because we're multipling $R_{nl}(0)$ by zero squared.

At $r=2a_0$ you should see that the bracketed linear factor is zero.