# Quantum Level Question

1. Jun 22, 2004

### Dlockwood

Using the hydrogen model, when the electron drops to the 'lowest' quantum level it emits photons in the dark blue; when it drops to the second lowest level it emits in the blue-green. When it is 'bumped up' to the second level it absorbs photons (assuming more than one atom) and when it is 'bumped up' to the third level it absorbs photons. What are the frequencies of absorbtion? Have they been identified or are they assumed by reason of emissions when the electrons drop to lower quantum levels? And what are all of these exact frequencies?

2. Jun 22, 2004

### jcsd

I don't know what frequency light is emitted when a bound electron moves up or down an energy level iin a Hydrogen atom.

the boundary conditions for a particle in a potential means that bound electron in a helium atom can only have certain discrete energies.From the conservation of energy to move up from one energy level to antother an electron must gain the difference in energy between these two energy levels.

3. Jun 22, 2004

### chroot

Staff Emeritus
Actually, the set of transitions from n > 1 to n = 1 are called the Lyman series, and are all in the ultraviolet. The set of transitions from n > 2 to n = 2 are the Balmer series, and they are all in the visible.

You can easily calculate the energy (and frequencies) of the photons produced by these transitions, which are the same as those required to stimulate the reverse transition.

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

- Warren

4. Jun 22, 2004

### jcsd

Why did I say "helium atom"?

5. Jun 24, 2004

### tavi_boada

Wouldn't it simply be the absolute value of the energy difference divided by h?

6. Jun 24, 2004

### Tom Mattson

Staff Emeritus
Yes, they can be derived--to varying degrees of accuracy--from the Bohr model, the Schrodinger equation, the Dirac equation and QED. It's amazing that even the humble Bohr model does a pretty good job of it.

Exact? That we can't do. For that we'd have to solve QED exactly, which can't be done. However, to get you started, the prediciton of the Bohr and of the Schrodinger equation (in the absence of any magnetic effects) is:

En=(-13.6 eV)/n2

where n=1,2,3,...

The frequency of a photon emitted when an electron is demoted from a state m to a state n (m>n) is (Em-En)/h. If the electron is promoted (m<n), then the frequency of the absorbed photon is the negative of that formula.

7. Jun 24, 2004

### Tom Mattson

Staff Emeritus
Yes, but that just changes the question from, "What are the exact photon frequencies?" to "What are the exact atomic energies?"

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