# Hydrogen Atom; energy state transitions

1. Jan 2, 2010

### hhhmortal

1. The problem statement, all variables and given/known data

A hydrogen atom is initially in the state n=3. Subsequently it falls to its ground state with the emission of a photon. If the photon energy is ћω , then what is the ground state energy?

3. The attempt at a solution

I tried using E= h².n²/8ma².

and got -9ћω/8

I wasn't to sure how to go about with this question.

2. Jan 3, 2010

### Dick

Uh, how did you get a negative ground state energy using that formula?

3. Jan 3, 2010

### hhhmortal

Not too sure. I think that formula would just give you the energy of the level, but doesnt the electron in that level require that amount of energy to be ionised, hence the negative sign.

4. Jan 3, 2010

### Dick

You are right about that. And in fact, I think your answer is right. On the other hand the formula you gave for E is wrong. Why do you try and write down a clear derivation of your result?

5. Jan 5, 2010

### hhhmortal

I'm not too sure, I think I just used that equation then I cancelled out certain terms and multiplied it by ћω. This is wrong, but I can't find the right way of doing it.

6. Jan 5, 2010

### Dick

Start by finding the correct formula. The formula for the energy levels has an n^2 in the denominator, not the numerator. How can you be 'not to sure' of how you got the answer???!