Is liquid helium composed of both ortho and para-helium?


by Creator
Tags: composed, helium, liquid, ortho, parahelium
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#1
Jun23-10, 10:25 PM
P: 534
In a batch of liquid helium (say < 8* K.) is there a mixture of both ortho and para-helium, or is it only ground state para-helium.?
I was under the impression that ortho was a meta-stable state which cannot decay to ground state Para-helium by radiative emission, but by meta-stable we are talking only a fraction of a second...right?

What am I missing here.?

Anyone?
..
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alxm
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Jun23-10, 10:49 PM
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IIRC, the singlet-triplet splitting is quite large, ~20 eV. So it should be entirely in the ground state, unless there are special conditions going on (they've made triplet-helium BECs).

It can decay radiatively, just not by a single-photon process.
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#3
Jun23-10, 11:18 PM
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Quote Quote by alxm View Post
IIRC, the singlet-triplet splitting is quite large, ~20 eV. So it should be entirely in the ground state, unless there are special conditions going on (they've made triplet-helium BECs).
Thanks alxm...
What special conditions are you referring to ? How do I create triplet state experimentally?

It can decay radiatively, just not by a single-photon process.
So if triplets decay after 1/10 second (is that about correct?) then by what process do they decay?

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Jun23-10, 11:51 PM
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Is liquid helium composed of both ortho and para-helium?


Quote Quote by Creator View Post
What special conditions are you referring to ?
Well I was thinking if you specifically generate triplet helium and then cool it in an optical trap or similar. Not exactly your usual state of affairs.

So if triplets decay after 1/10 second (is that about correct?) then by what process do they decay?
This is a 'forbidden' transition; It can only occur (in an isolated atom) due to two-photon processes, which one can think of as a decay to a 'virtual' level in-between the two states. For Helium it's not symmetric; one of the photons carries more energy than the other, so you end up with two peaks, one at ~70 nm and one at ~2400. (See e.g. this paper, if you want some details)

1/10 of a second.. I'm not sure. An isolated 3He atom would probably have a lifetime orders of magnitude longer. But with gaseous or liquid helium, in a container, etc, you have many more interactions that can go on and assist the process. (For instance formation of a He2 molecule in the [tex]^3\Sigma_u^+[/tex] state.)
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#5
Jun25-10, 12:45 PM
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Quote Quote by alxm View Post
Well I was thinking if you specifically generate triplet helium and then ..
That was my other main concern: By what method would you "specifically generate triplet helium"??(not BEC)


This is a 'forbidden' transition; It can only occur (in an isolated atom) due to two-photon processes, which one can think of as a decay to a 'virtual' level in-between the two states. For Helium it's not symmetric; one of the photons carries more energy than the other, so you end up with two peaks, one at ~70 nm and one at ~2400. (See e.g. this paper, if you want some details)
Thanks for the G W Drake link. I'm not too sure I understand it all....Of course, Selection rules prohibit single photon transitiion .... so it goes two photon route...OK, but this rate is extremely low, right?
I have read elsewhere that the most probable decay is thru 'collisional' process, I am not sure what that refers to ...any info there?

GW Drake has a later article in which it appears as if he is saying that there is a MORE probable Magnetic dipole transition to ground state decay...am I reading that right?
See here:

http://cos.cumt.edu.cn/jpkc/dxwl/zl/...atomic/099.pdf


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alxm
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#6
Jun25-10, 02:36 PM
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Quote Quote by Creator View Post
That was my other main concern: By what method would you "specifically generate triplet helium"??
I haven't researched it, but off the top-of-my head you could simply irradiate the helium with a wavelength that excited it from the singlet ground state to a state that was able to decay to the triplet state. Given that that state has a relatively long lifetime, this should give you all the triplet helium you want, given that the radiation is sufficiently intense.

.OK, but this rate is extremely low, right? I have read elsewhere that the most probable decay is thru 'collisional' process, I am not sure what that refers to ...any info there?
Yes, that's the most probably situation for a non-isolated helium, and in practice the most common decay method. The easiest and most common is simply that two triplet-helium atoms collide and exchange an electron (or viewed another way: flip each other's spins). That process is fairly straightforward.

GW Drake has a later article in which it appears as if he is saying that there is a MORE probable Magnetic dipole transition to ground state decay...am I reading that right
It appears so. Guess I was wrong. Given that it appears Breit and Teller also thought the two-photon process was more likely, at least I'm in excellent company!
Both forms of decay are certainly many orders-of-magnitude less significant than the collision route, in a 'normal' environment. This is all mostly of astrophysical interest, since space is full of helium atoms in high vacuum.
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#7
Jul1-10, 10:08 PM
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Quote Quote by alxm View Post
Yes, that's the most probably situation for a non-isolated helium, and in practice the most common decay method. The easiest and most common is simply that two triplet-helium atoms collide and exchange an electron (or viewed another way: flip each other's spins). That process is fairly straightforward.
Thanks again, Alxm. and sorry for the late response.

The collisional process is interesting...apparently, collision with a container wall can also do the trick.
1st question: At what temperature is there enough kinetic energy to spin flip each He triptlet state? Probably somewhere there is a temperature dependent rate formula?
2. Can collisional spin flip be by mecahnical means, (ex,rotation of the fluid) ?


It appears so. Guess I was wrong. Given that it appears Breit and Teller also thought the two-photon process was more likely, at least I'm in excellent company!
Both forms of decay are certainly many orders-of-magnitude less significant than the collision route, in a 'normal' environment. This is all mostly of astrophysical interest, since space is full of helium atoms in high vacuum.

After further reasearch it appears as though I was wrong about the 1/10 sec. excited lifetime.
The He(2S^3) state lifetime is about 8000 seconds!! That's a quantum eternity, and surprising...probably the longest"Meta-stable" state around.....but makes sense knowing the fact that it is radiatively forbidden transition to ground.

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