Which photon has the higher frequency?

1. Apr 2, 2005

smokie

Chemistry is not my strong subject...

A hydrogen atom undergoes a transition from the state n=3 to n=2 and then another transition from n=2 to n=1. Two photons are created due to these processes. Which photon has the higher frequency?

AND What is meant by 'n = #' ?

2. Apr 2, 2005

SpaceTiger

Staff Emeritus
Check out this thread, keeping in mind that $$E=h\nu$$. The principal quantum number, "n", is a property of a quantum mechanical state with certain energy. States with different "n" have different energies.

3. Apr 6, 2005

smokie

I didn't understand the discussion on that site very well. The course I'm in is a physics course and they threw in some chemistry, however, we don't have a textbook, so my resource is mainly the internet...

States with different 'n' values have different energies, but do the energies from n = 1, 2, 3 differ by a constant value [i.e. 1, 3, 5] or not?

4. Apr 6, 2005

SpaceTiger

Staff Emeritus
Actually, the energy of a state is proportional to the square of the inverse of the principal quantum number, "n". So, if we choose units in which the energy of the n=1 state is -1 (the total energy of a bound state is always negative), then the n=2,3, and 4 states will have energies of -1/4, -1/9, and -1/16, respectively. An energy greater than 0 will mean that the electron is no longer bound to the nucleus.

5. Apr 6, 2005

smokie

ok got that...so then obviously an electron moving from energy state -1/9 to -1/4, it would give off more energy than one moving from -1/4 to -1, correct?

6. Apr 6, 2005

SpaceTiger

Staff Emeritus
You might want to try that subtraction again.

7. Apr 6, 2005

smokie

Well I can see what I did wrong but doesn't an electron moving to a state 'closer' to the nucleus give off energy?

8. Apr 6, 2005

SpaceTiger

Staff Emeritus
If by "closer" you mean more tightly bound to the nucleus, then yes. The most tightly bound state is n=1.

9. Apr 6, 2005

smokie

Yes, what I mean is that the electrons are more tightly bound to the nucleus. So, how should I do my subtraction? I should do -1/4 - -1/9 ? <-- Negative number would be required for energy released... ?

Last edited: Apr 6, 2005
10. Apr 6, 2005

SpaceTiger

Staff Emeritus
When it moves from state n=3 to n=2, does the electron's energy increase or decrease (based on what I already said)? If energy is conserved, what does this mean for the photon that's emitted?

11. Apr 6, 2005

smokie

When moving from state n = 3 to n = 2, the electron's energy decreases... it loses some of the energy to the escaping photon...that's what I'm understanding..

12. Apr 6, 2005

SpaceTiger

Staff Emeritus
That's right. So, to conserve energy, the atom must emit a photon of light!

13. Apr 6, 2005

smokie

Exactly, soo, but then isn't my subtraction correct?

The only thing I'm seeing wrong with my subtraction is that for 'energy' I should have a negative answer...

14. Apr 6, 2005

SpaceTiger

Staff Emeritus
In our units, the energy change from n=3 to n=2 is (-1/9)-(-1/4)=5/36. This is the energy that's released as light. It's the change in energy of the electron that's negative: (-1/4)-(-1/9)=-5/36.

15. Apr 6, 2005

smokie

Ahh, I see what you mean. E2 - E1 = energy released as light, while E1 - E2 = change in energy of the electron.

16. Apr 6, 2005

smokie

That would mean that more energy is released from n = 2 to n = 1, than is from n = 3 to n = 2, would it not?