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The wavefunction for an electron within a hydrogen atom in the $2s$ state has the following wavefunction:

$$\psi(r,\phi,\theta)=\psi(r)=\frac{1}{2\sqrt{\pi}}\left(2-\frac{r}{a_0}\right)\frac{e^{-r/2a_0}}{(2a_0)^{3/2}}$$

However, at $r=0$,

$$\psi^*\psi\left.\right|_{r=0}=\left(\frac{1}{2\sqrt{\pi}}\left(2-\frac{0}{a_0}\right)\frac{e^{-0/2a_0}}{(2a_0)^{3/2}}\right)^2=\frac{1}{\pi(2a_0)^{3}}$$

I don't understand how this should be possible. My answer doesn't match logic and it doesn't match graphs that I find online. (every graph I see goes to zero at $r=0$)

However, this does seem to really be the wavefunction for the 2s state.

Where have I gone wrong?

Arturo don Juan
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1 Answers1

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The graph shows the probability of finding the electron between the distances $r$ and $r + dr$. This probability is given by:

$$ P = \psi^* \psi dV $$

where $dV$ is the volume element:

$$ dV = 4\pi r^2 dr $$

So we get the probability:

$$ P(r,r+dr) = \psi^* \psi 4\pi r^2 dr $$

and therefore when $r = 0$ the probability $P = 0$. It isn't that the wavefunction goes to zero at $r = 0$, but that the size of the volume element goes to zero.

John Rennie
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