# Genesis of the H-α

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## Main Question or Discussion Point

Hi,
I am trying to learn the origins of the Hydrogen spectral series.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
I found that H-α has wavelength 656.28525 nm that corresponds to 1.8892 eV (the PE difference from energy levels 3 and 23.4 - 1.511)
but the electron needs a similar amount of KE, falling from r = 4.76 to 2.12 x 10^-7 cm.) and increasing its speed
- where does it get it from?
- why the KE gain is equal to the E of emitted photon?
- can we calculate/ measure the time it takes to move from one level to the other?
- does the electron keep circling while falling?
Thanks

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1- where does it get it from?
2- why the KE gain is equal to the E of emitted photon?
3- can we calculate/ measure the time it takes to move from one level to the other?
4- does the electron keep circling while falling?
Thanks
1: As you said, the energy comes from the potential energy
2: That's a consequence of the virial theorem
3: No. The electron doesn't follow a classic trajectory so the question doesn't make sense
4: No. The electron doesn't follow a classic trajectory so the question doesn't make sense

1 person
Gold Member
1: As you said, the energy comes from the potential energy
Thanks, dauto, in the link I quoted the difference of energy is only 1.88 eV, that is enough for the photon, have I misread?

I found that H-α has wavelength 656.28525 nm that corresponds to 1.8892 eV (the PE difference from energy levels 3 and 23.4 - 1.511)
but the electron needs a similar amount of KE, falling from r = 4.76 to 2.12 x 10^-7 cm.) and increasing its speed
These energy levels represent the energy (total) not just PE.

1 person
Thanks, dauto, in the link I quoted the difference of energy is only 1.88 eV, that is enough for the photon, have I misread?
1.89 eV is the energy difference between the levels which includes both the kinetic and potential energy changes. The potential energy change is twice as big (with opposite sign) as the change in kinetic energy - because of the virial theorem as I said.

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1 person
Gold Member
The potential energy change is twice as big (with opposite sign) as the change in kinetic energy - because of the virial theorem as I said.
Thanks for your help, I tried this http://en.wikipedia.org/wiki/Virial_theorem
but it's too hard for me , can you explain how it works? is it that the electron can use up only half of the energy?

Could you show me practically how to calculate the energy the electron gains from H3 to H2?
Fcm = 2.3 * 10-19
r3 = 9 * rbohr
r2 = 4 * rbohr
the drop is 5r = 2.645 * 10-8cm

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Thanks for your help, I tried this http://en.wikipedia.org/wiki/Virial_theorem
but it's too hard for me , can you explain how it works? is it that the electron can use up only half of the energy?

Could you show me practically how to calculate the energy the electron gains from H3 to H2?
Fcm = 2.3 * 10-19
r3 = 9 * rbohr
r2 = 4 * rbohr
the drop is 5r = 2.645 * 10-8cm
From the wiki page you quoted:

$2 \langle T \rangle = n \langle V_\text{TOT} \rangle,$

where n is taken from the potential V(r) = αrn

for the case of an electromagnetic force, the potential is given by V(r) = αr-1, so clearly we have n = -1. plugging that back into the virial equation we get

$2 \langle T \rangle = - \langle V_\text{TOT} \rangle.$,

in other words the average potential energy is twice as big (with opposite sign) to the average kinetic energy.

1 person
Thanks for your help, I tried this http://en.wikipedia.org/wiki/Virial_theorem
but it's too hard for me , can you explain how it works? is it that the electron can use up only half of the energy?

Could you show me practically how to calculate the energy the electron gains from H3 to H2?
Fcm = 2.3 * 10-19
r3 = 9 * rbohr
r2 = 4 * rbohr
the drop is 5r = 2.645 * 10-8cm
The electron does not gain but lose energy when it goes from level 3 to level 2.
The change in energy is ΔE=E2-E3=-3.4eV -(-1.5eV)=-1.9eV.
So it loses 1.9 eV.

Gold Member
The electron does not gain but lose energy when it goes from level 3 to level 2.
The change in energy is ΔE=E2-E3=-3.4eV -(-1.5eV)=-1.9eV.
So it loses 1.9 eV.
It loses PE but gains KE as it increases its speed from 1/3 to 1/2 of 2.188*108c/s, correct?

Gold Member
in other words the average potential energy is twice as big (with opposite sign) to the average kinetic energy.
Thanks, dauto,
To what KE are you referring?
the speed at
H3 is 2.18*108/3, KE3 = 13.6/9 = 1.51 eV
H2 is 2.18*108/2, KE2 = 3.4 eV
ΔE= 13.6*5/36 =1.89 eV

f= 2.3*10-23 dyn, right?
the formula to find ΔPE is f *5/36r =3.78 eV, right?

If you give an electron in H2 3.78 eV energy, will it emit a 1.89 eV photon?

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jtbell
Mentor
It loses PE but gains KE as it increases its speed from 1/3 to 1/2 of 2.188*108c/s, correct?
Yes. The amount of PE lost is greater than the amount of KE gained, so the total E decreases.

Thanks, dauto,

If you give an electron in H2 3.78 eV energy, will it emit a 1.89 eV photon?
No, you don't have to give it anything.
It will spontaneously "go" from level 3 to 2 and emit a photon.
You need to give him energy to go the other way.

I am still trying to understand what are you after with all this.
The change in potential energy is negative (it decreases). You keep moving in circles ignoring the things already discussed. So it's not 3.78 eV but -3.78 eV.

Gold Member
No, you don't have to give it anything.
It will spontaneously "go" from level 3 to 2 and emit a photon.
You need to give him energy to go the other way.
I am still trying to understand what are you after with all this.
.
That is what I am asking: an e is in a level (say H2), if you give energy 1.89 eV, will it jump to level 3?, if you give him more (must it be 2*1.89 ?), will it emit a photon?

That is what I am asking: an e is in a level (say H2), if you give energy 1.89 eV, will it jump to level 3?, if you give him more (must it be 2*1.89 ?), will it emit a photon?
Yes to first. No to second. Unless you are thinking about stimulated emission, as in laser. In which case the photon goes back to level 2 in the process and will be two photons at the end. But I doubt this is what you have in mind.
It seems you have some fundamental confusion or misunderstanding.
Splitting the energy into potential and kinetic is not so relevant. It's simply that transitions of electrons between energy levels (total energy) are accompanied by emission or absorption of photons.
If it goes "up" it needs to absorb a photon. If it goes "down" it emits a photon.
Trying to understand these in terms of classical mechanics can take you to some point but not farther.

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Gold Member
But I doubt this is what you have in mind.
...
If it goes "up" it needs to absorb a photon. If it goes "down" it emits a photon.
.
I am a student , I'm trying to understand the behaviour of electrons.
When it goes down it gets rid of the surplus PE ,
when it goes back up from 2 to 3 it needs the same amount of energy 1.89 eV, if you give it more, it must absorb absorb all you give, then to stay at level 3, it should get rid of surplus energy, is this correct?

DrClaude
Mentor
When it goes down it gets rid of the surplus PE ,
As nasu said, it is best to stop thinking in terms of PE and KE, and just think of the total energy of the electron.

when it goes back up from 2 to 3 it needs the same amount of energy 1.89 eV, if you give it more, it must absorb absorb all you give, then to stay at level 3, it should get rid of surplus energy, is this correct?
This is not correct. The energy must match exactly. If the photon carries an amount of energy that is not equivalent to the difference between the current energy level of the electron and another accessible level, it will not be absorbed.

This is why an absorption spectrum is made up of discrete line, corresponding to the possible transitions.

Gold Member
The energy must match exactly. If the photon carries an amount of energy that is not equivalent to the difference between the current energy level of the electron and another accessible level, it will not be absorbed.
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That is what is hard to understand, if the energy is in excess it can refuse the whole energy, or I got it wrong again?

DrClaude
Mentor
That is what is hard to understand, if the energy is in excess it can refuse the whole energy, or I got it wrong again?
It doesn't "refuse" the energy. There is just no significant interaction between the photon and the atom if the energies don't match. Welcome to quantum mechanics!

Actually, you see the same thing in classical systems. If you have a system that has a particular oscillation frequency, you can only excite it if you provide energy at that same frequency. Just think of a radio.

1 person
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Actually, you see the same thing in classical systems. If you have a system that has a particular oscillation frequency, you can only excite it if you provide energy at that same frequency. Just think of a radio.
Yes also a string as only one frequency of resonation and knows it very well.
But here is different, an electron in a level cannot "know" what energy it takes to go to another level. And , after all, I thought, it cannot in any way resist if you apply an electric or magnetic field, or can it?
When you give it k energy it moves, i you give it k+ it should react even faster, I thought

DrClaude
Mentor
Yes also a string as only one frequency of resonation and knows it very well.
The string doesn't known anything. There is an external time-periodic force which it reacts to, and depending on the "natural" response frequency of the string, it gets excietd or not, which means that it keeps absorbing the energy from the outside stimulation, rather than it averaging to zero.

But here is different, an electron in a level cannot "know" what energy it takes to go to another level.
Thinking classically, you can see it the same way. If the frequency of the photon matches what is needed for the electron to change orbitals, then it can be absorbed.

By the way, talking just about the elctron is actually incorrect. It is the atom itself that absorbs the energy. It just happens that, the electron having such a small mass compared to the nucleus, it is the electron that appears to change state. So you can see the dipole made up of the nucleus and the elctron as a little antenna.

And , after all, I thought, it cannot in any way resist if you apply an electric or magnetic field, or can it?
Instantaneously, in the presence of an external electric or magnetic field, yes, the electron will be affected. But what is relevant here is the before and after picture, in which the electron stays in the same orbital if the energy of the photon doesn't match.

When you give it k energy it moves, i you give it k+ it should react even faster, I thought
I think that you are trying to get too much out of a classical picture. The electron and the photon are ruled by the laws of quantum mechanics, and there is simply no mechanism that will allow the atom to simultaneously absorb a photon of on frequency and emit one of another frequency at the same time. (Note that I am talking here about the interaction with a single photon. There are other processes that can appear when you have lots of photons at the same time, like when using a laser.)

1 person
Gold Member
Instantaneously, in the presence of an external electric or magnetic field, yes, the electron will be affected. But what is relevant here is the before and after picture, in which the electron stays in the same orbital if the energy of the photon doesn't match.
That is what I was saying, it must be affected, then it falls back, and emits a photon, right?

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DrClaude
Mentor
That is what I was saying, it must be affected, then it falls back, and emit a photon, right?
In the presence of an external electric or magnetic field, the orbital of the electron will be perturbed by the presence of the field. If the field is weak, this orbital is a slight modification of the corresponding orbital in the absence of the external field. But you rcan't eally consider that as the electron being excited. The only exception here is if that field is extremely strong, but I don't think this is what you are thinking of here.

If it is a single photon passing by, you can basically neglect the effect of the field of that photon: the electron is unaffected by the presence of the photon. The only exception is if the photon is of the right frequency, in which case there is some probability (but not certainty) that the energy of the photon will be absorbed, and the electron will end up in an excited state.

Until you learn about quantum mechanics, you should keep it simple: if the photon is of the right frequency, it gets absorbed. Otherwise, it is as if it wasn't there. Just like a radio: if an EM field comes by a radio but the field is not the right frequency, you will not here any music, period.

1 person
Gold Member
The only exception here is if that field is extremely strong, but I don't think this is what you are thinking of here.
No, I was just thinking of energy slightly greater or double than needed.
Thanks, you've been very helpful, I'll digest that and maybe come back with further doubts.