# Energy levels

1. Oct 20, 2009

### Ylle

Hi...

I'm reading about "Particle in a box", and I really can't seem to figure out what "Energy Levels" actually means. The thing in my book is a bit slacky, so I'm not quite sure that I understand. So could anyone maybe explain to me what the "Energy Levels" in "Particle in a box" is ?
Doesn't have to be anything fancy, just so I have a clue :)

Regards.

2. Oct 20, 2009

### Anti-Meson

The energy levels are from the time (dependent or independent - depending how simple you want the answer) Schrodinger equation.

i.e. The time-independent Schrodinger equation in one dimension:

Hψ = Eψ

Where ψ is the wave function of the particle, H is the Hamiltonian operator and E is the energy of the particle (its kinetic and potential).

Forgot to add: E is an eigenvalue for the eigenfunction ψ. If you don't know what eigenvalues/functions are just ignore that part.

Last edited: Oct 20, 2009
3. Oct 20, 2009

### Ygggdrasil

We normally think of energy as a continuous variable. For example, a baseball move with 10 J of kinetic energy, 20 J of kinetic energy or any value in between. In quantum mechanics, energy is discrete (quantized): that is, a particle can have only certain values of energy and intermediate values of energy are forbidden. If we were to apply this property to the baseball, it would mean that the baseball could, for example, have 10J of kinetic energy or 20J of kinetic energy, but you could never find it moving with 15 J of kinetic energy (or 11 J, 12 J, etc).

This concept is strange, but the particle in a box example demonstrates why this is the case. Only waves whose wavelength fits a particular condition (the boundary conditions) can exist inside of the box (or else the waves destructively interfere with themselves and disappear). This set of allowed waves constrains the allowed energies of the particle to the energies associated with those waves. Particles with intermediate energy have waves which do not fit into the box and therefore are forbidden.

These allowed energies are the energy levels. A good analogy here is a ramp (allows any value for gravitational potential energy) versus stairs (which give discrete energy levels and forbid intermediate energies).

4. Oct 21, 2009

### Ylle

So you have to see energy levels as the only energy allowed in that precise atom fx. ?
So energy levels are levels that are allowed in certain things, but not necessarily has to be there ?

I mean, the hydrogen atom energy levels, in the Bohr Model, is given by:

$${{E}_{n}}=-\frac{1}{{{\varepsilon }_{0}}}\frac{m{{e}^{4}}}{8{{n}^{2}}{{h}^{2}}}$$,
where n is the quantum number.

So I can make infinite energy levels in the hydrogen atom just by choosing n = 1,2,3,... ?

Or does it have some sort of limitation to it ?

Last edited: Oct 21, 2009
5. Oct 21, 2009

### Bob_for_short

Energy levels belong to the so called stationary states - the states that do not change with time. They are just proper frequensies of systems. Intermediary energies are also allowed - they are obtained as superpositions of discrete states but they are not stationary - they evolve.

Hydrogen may be in different stationary states or in a superposition of them.

6. Oct 21, 2009

### Staff: Mentor

In principle, you can use any value of n, as large as you like. But as n becomes very large, the energy becomes very close to zero, which means the electron is very weakly bound to the proton, and it's very easy to "knock away" by collision with another atom or particle. So in practice you can't produce and maintain hydrogen atoms with very large values of n. I don't know what the practical upper limit is.

Some physicists specifically study such atoms with large values of n, because they give us some insight into the "transition" between quantum and classical physics.

7. Oct 21, 2009

### Ylle

But does that mean that there IS energy levels in fx. n = 100, or just that it is possible ?
'Cause the hydrogen atom only got 1 electron, right ? So shouldn't that be in the most possible energy level (n = 1) all the time ? Or is it when we put energy into the atom, that we can excite it to a different energy level (For n = 1,2,3,...) ?

8. Oct 21, 2009

### Bob_for_short

Yes, such energy level exist really and sometimes such states are very stable (see Rydberg atoms).

One can excite an atom to high n indeed, in collisions with other particles.

You know, in plasma where the electrons are detached from protons, there are processes of recombination that occur via states with high n. Gradually electron orbit decreases, the quantum number n decreases too (plasma's cooling down).

Last edited: Oct 21, 2009
9. Oct 21, 2009

### G01

The energy level is there. It is just that for states with really high n, that energy is very close to zero, meaning the electron can be knocked free from the atom very easily if it is in those energy states.

Normally, yes, the electron will be in the ground state (n=1). However, you can excite the electron to a higher energy level by letting it absorb a photon with an energy equal to the difference between the levels.

10. Oct 21, 2009

### Ylle

Ahhh, I see...

But is there any way you can determine a stable state of energy level ?
Isn't an excited electron always unstable ? Wouldn't it just go back to it's normal energy level, and then emmit a photon with energy of whatever I've put into it to excite it ?

And just another one.
You have the energy level for hydrogen atom given by:

$${{E}_{n}}=-\frac{1}{{{\varepsilon }_{0}}}\frac{m{{e}^{4}}}{8{{n}^{2}}{{h}^{2}}}$$

If n = 1, you get 13.6 eV, that should be the energy required to completely remove the electron from the hydrogen atom, right ?
So if I put n = 2, I get the energy required to excite to energy level 2 ?
And for n = 3 it gets even smaller and so on. So it's easier to excite an electron to n = 100 than n = 1 ?
Or does it just mean, that IF an electron IS in energy level 100 fx., then it would take En (for n = 100) to remove that electron from the atom ?

11. Oct 21, 2009

### Bob_for_short

These levels are just like stairs: to excite n=2 you have to supply at least E2-E1. It is the level difference that is necessary to make a transition.

Some states with high n are metastable (Rydberg atoms). They are very symmetric and hard to change spontaneously. Dipole transitions are suppressed. You have to know quantum mechanics to understand that.

Last edited: Oct 21, 2009
12. Oct 21, 2009

### Ylle

I see :)

So, what if I have a difference of 100 to n = 2 and 200 to n = 3 (Random numbers), and I put in 150 into the electron. Would that excite it to level 2, or does it have to be exact ? If not, what happens to rest of the energy (The 50) ?

13. Oct 21, 2009

### Bob_for_short

In order to make a transition from low n1 to higher n2 you have to supply at least the energy difference En2 - En1. If you have more, the excess of energy can be spent on the kinetic energy of the atom as a whole.

For example, you hit an atom with a fast electron. As a result you can obtain an excited and moving atom with any n' that is allowed with the energy conservation law, including an ionized atom.

Excitation part of the energy can be considered as a change of internal energy of the system (inelastic projectile-atom collision).

Last edited: Oct 21, 2009
14. Oct 21, 2009

### Ylle

That makes sense :)

Thank you very much.

15. Oct 21, 2009

### Bob S

The table on hydrogen atomic level lifetimes up to 6f in Bethe and Salpeter "QM of 1 and 2 Electron Atoms" (page 266) shows that the longest calculated lifetime is the 6s state, with a lifetime of 570 namnoseconds. None, other than the 1s state, is stable.
Bob S

16. Oct 21, 2009

### Bob_for_short

Yes, metastable states are s-states with high n.

17. Oct 21, 2009

### mikeph

My explanation would be that there are discrete solutions to the time-independent Schroedinger equation for a particle in an infinite potential well, and associated with these discrete wavefunctions come discrete energy eigenvalues. The more complicated potential well due to a fixed proton is a similar story- there are discrete wavefunction solutions, so there are discrete energy levels.

What I was wondering was with regards to the stability of states- we are solving the time-independent Schroedinger equation, so how can we do stability analysis? Do we just assume that a higher energy state 'prefers' to go into a lower energy one like in school? Is any perturbation theory done to find that n=2 will decay into an empty n=1 given enough time? And by metastable, do you mean, will stay in that position with zero perturbation? Or the perturbations will not grow?

18. Oct 21, 2009

### Bob_for_short

In fact we are speaking here of charged particles. Any charged particle is coupled with the quantized electromagnetic field, it never come alone. So purely stationary states are possible with neglecting the quantized EMF influence.

As soon as one wants to calculate the rate of transition from, say, n=2 to n=1, one has to take into account the quantized EMF in the Hamiltonian. It is a perturbation term and it allows redistribution of energy between the atomic end filed degrees of freedom. So the transition happens with emitting a photon.

In dense medium there is interaction with the neighboring atoms so there may be transfer of energy to another atom without photon emission.

19. Oct 21, 2009

### Bob S

From Bob S
The table on hydrogen atomic level lifetimes up to 6f in Bethe and Salpeter "QM of 1 and 2 Electron Atoms" (page 266) shows that the longest calculated lifetime is the 6s state, with a lifetime of 570 namnoseconds. None, other than the 1s state, is stable.
Only the 2s state in hydrogen is metastable. All other transitions, including s states for n>2, have allowed transtions to states with lower n.
Bob S