Which photon has the higher frequency?

[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 = #' ?

 

[SpaceTiger]

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.

 

[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?

 

[SpaceTiger]

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?

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.

 

[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?

 

[SpaceTiger]

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?

You might want to try that subtraction again.

 

[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?

 

[SpaceTiger]

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

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

 

[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... ?

 

[SpaceTiger]

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... ?

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?

 

[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..

 

[SpaceTiger]

When moving from state n = 3 to n = 2, the electron's energy decreases... that's what I'm understanding..

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

 

[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...

 

[SpaceTiger]

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.

 

[smokie]

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

 

[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?