I Wavefunction of ionized hydrogen electron

[Physicsman88]

TL;DR

When you solve the schrodinger equation and get the wavefunctions for hydrogen is there a point where the quantum number becomes so high the electron gets ejected and breaks free. If so what would the wave function look like

At what point does the electron become ionized in the hydrogen atom solution

 

[DrClaude]

There is an infinite number of bound states, so no, there is no "point where the quantum number becomes so high the electron gets ejected and breaks free."

That said, even as $n \rightarrow \infty$ the energy remains finite, so it is possible to have an electron no longer bound to the nucleus. Its wave function then looks like a scattering state, which is basically an oscillating sine wave, except close to the nucleus.

 

[Physicsman88]

There is an infinite number of bound states, so no, there is no "point where the quantum number becomes so high the electron gets ejected and breaks free."

That said, even as $n \rightarrow \infty$ the energy remains finite, so it is possible to have an electron no longer bound to the nucleus. Its wave function then looks like a scattering state, which is basically an oscillating sine wave, except close to the nucleus.

What happens to the angular component of the wavefunction when the electron gets ionized.

 

[Physicsman88]

There is an infinite number of bound states, so no, there is no "point where the quantum number becomes so high the electron gets ejected and breaks free."

That said, even as $n \rightarrow \infty$ the energy remains finite, so it is possible to have an electron no longer bound to the nucleus. Its wave function then looks like a scattering state, which is basically an oscillating sine wave, except close to the nucleus.

How do you show that as n goes to infinity The radial wavefunction becomes a sine wave

 

[DrClaude]

What happens to the angular component of the wavefunction when the electron gets ionized.

That depends on how ionization takes place.

How do you show that as n goes to infinity The radial wavefunction becomes a sine wave

That's not what happens. For E < 13.6 eV, you have an infinite number of bound states with the wave function a product of a Laguerre polynomial and a decaying exponential. For E > 13.6 eV, the solutions are sines as $r \rightarrow \infty$, like what is obtained for a (finite) square potential, as found in most QM textbooks.

 

[Physicsman88]

How do you mathematically show that the solutions are sines for the radial wavefunction when E>13.6 eV like is there a proof for it

 

[Vanadium 50]

How do you mathematically show that the solutions are sines for the radial wavefunction when E>13.6 eV like is there a proof for it

Since they aren't, you don't. They are sines for unbound particles in Cartesian coordinates.

 

[Physicsman88]

Since they aren't, you don't. They are sines for unbound particles in Cartesian coordinates.

So if the unbounded radial wavefunctions arent sine waves what are they then

 

[Vanadium 50]

So if the unbounded radial wavefunctions arent sine waves what are they then

My slaves! Solve this equation for me!

Are you going to put in any effort at all? Seriously - if careful reading of replies is too much effort, and Googlinmg "wave equation in spherical coordinates" is too much effort, how do you expect to make any progress? We can't simply pour information into your head.

For an l=0 case, the wavefunction is of the form sin(kr-wt)/r.

 

[HomogenousCow]

The unbound hydrogen states are briefly mentioned on page 231 of these notes.