Ground state of hydrogen atom

Question

My interpretation: When we have no angular momentum, the potential well looks like this, my question is: How do you find the point where the wavefunction penetrates its classical forbidden region, i.e. quantum tunnelling effect? The answer is $r>2a_0$ where $a_0$ is the Bohr's radius.

My answer: It seems you have to equate the Coulomb potential with the energy of the ground state of the hydrogen atom.

And what would the wavefunction intersect? Does it pass through $(0,0)$? Can we view the infinite deep well, as a infinite barrier, by imagining we shift the zero reference level down to infinity?

I'm quite confused by the idea of having negative energy. Does it mean I need infinite energy to get out of the well? Just like an infinite square well?

Answer 1

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

Answer 2

Before turning to your problem in the next paragraph, let's consider throwing a ball up in the air. Classically, you can compute its potential and hence kinetic energy at any height, and at forbidden heights you'd get a negative value for the latter, which obviously can't match $mv^2/2$. But quantum mechanics doesn't mind; you just change the sign of an eigenvalue of $\nabla^2$, e.g. changing from plane waves to an exponentially decaying wavefunction. That's why quantum tunnelling is unlikely to take a particle far into the classically forbidden region (where "far" is measured by how positive the eigenvalue of $\nabla^2$ becomes).

Now to your problem. Classically, the electron has potential energy $-\alpha c\hbar/r$. Compare that to your formula $E=-mc^2\alpha^2/2$ when $n=1$; the critical $r$ is $2\hbar/ (\alpha mc)=2a_0$. If $r>2a_0$, the potential energy exceeds the eigenenergy, so the kinetic energy is negative; that's why it's a classically forbidden region.

Answer 3

For this model recall that the electron orbitsheld in place by a coulomb like force. $$ \frac{m_ev^2}{r}=\frac{4\pi\epsilon_0e^2}{r^2} $$ Now the total energy is given by the (coloumb like) potential and kinetic energy $$E=U+K$$ Use the first equation to get rid of the $v$ terms and solve for the ground state. See wikipedia for more details.

For your other questions:

The Bohr model is semi-classical in that are assumed to orbit the nucleus in a circular orbit at fixed radii. They don't pass through the origin. I'm not even sure it really makes sense to talk about their wave-functions.

It doesn't require infinite energy to escape the well as there is a minimum (finite) potential at the ground state.

This is the conventional place to set the a free unbound particle is zero energy. You'll find the same convention for gravitational potential.

Answer 4

I have discussed this problem with my friends, and he told me that it doesn't make sense to look at this problem classically, as the problem is inherently quantum. the wavefunction is everywhere around the potential. He said at the point of infinite -ve potential, we could have viewed the electron as having infinite energy, but it doesn't make any sense after all.