The h Bar

7:02 PM

booya

0

There is now some overlap between what we can do using optical elements, and the sort of photon energies that correspond to nuclear excitations. In particular, as mentioned in e.g. this paper, the 229Th nucleus has a low-lying excited state at energy $E=7.8\pm0.5\:\mathrm{eV}$, and this is low en...

rob

@EmilioPisanty Very nice. You didn't even get distracted by halfnium.

Emilio Pisanty

@rob wait, what?

what's hafnium gotta do with stuff?

rob

@EmilioPisanty Halfnium-178 has a electromagnetic excited state which is metastable with lifetime 30 years.

Emilio Pisanty

@rob huh

at what energy?

rob

There used to be proposals to chemically separate a bunch of this isomer to see whether the relaxation of one nucleus would stimulate emission on its neighbors.

Emilio Pisanty

7:12 PM

@rob huh

rob

Difference in mass excess between isomer and ground state is 2.4 MeV. But the isomer has $J^P = 16^+$. It's slow because the transision requires a lot of angular momentum.

Emilio Pisanty

yeah, that'd be pretty neat

@rob so that'd make it hard to produce the isomer?

AccidentalFourierTransform

@ACuriousMind it wasnt me this time. Someone else accepted the flag, I didnt even have time to see it :P

rob

There's an $8^-$ state at about half the energy, lifetime four seconds, so the decay process is emission of hexadecapole photons.

AccidentalFourierTransform

it was like, an instantaneous suspension lol

Emilio Pisanty

7:15 PM

@rob so what would be the relevant observable?

rob

@EmilioPisanty I think that if you put neutrons on Hf-177, some fraction of the new Hf-178 gets stuck in the isomeric states.

Emilio Pisanty

big cascades of photons instead of single ones?

@rob I guess the question was why is $J=16$ bad

also, if you've got references handy they'd be 'ppreciated

rob

@EmilioPisanty Probability of a transition decreases as $\Delta J$ increases. Electric dipole transitions are fastest; magnetic dipole and electric quadrupole are competitive with each other, etc.

Emilio Pisanty

@rob yeah, OK, so it's a high-$J$-makes-production-harder argument

ooooh, wait

or it's about the transition back?

rob

@EmilioPisanty Starting point: en.wikipedia.org/wiki/Hafnium_controversy

@EmilioPisanty Yes, you get ground-state and isomeric halfnium out of your reactor --- you don't pump it

Emilio Pisanty

7:20 PM

(Aside: @Qmechanic, I changed the redaction on the latest meta post to just [user]. Language along the lines of 'X-who-shall-not-be-named' has some pretty weird resonances for folks born '85 and later.)

@rob ok, got it

so the high $J$ makes the stimulated emission harder, then?

'cause you'd need to use the hexadecapole component of the incoming photon to stimulate the transition, I'd guess

rob

@EmilioPisanty That's the idea.

Emilio Pisanty

@rob so, what's the wavelength-to-system-size ratio there?

rob

That's what I thought of when I saw that question, but I probably won't have time to answer it.

@EmilioPisanty Hmmm, good question.

Emilio Pisanty

@rob thanks, btw, that's plenty to start a lit search

> The activity is in a cascade of penetrating gamma rays, the most energetic of which is 0.574 MeV

wait

cascade?

that'd make the hexadecapole argument funky

rob

@EmilioPisanty I guess it's not a single transition. I haven't looked at a level diagram.

@EmilioPisanty See if this link works for you: nndc.bnl.gov/useroutput/…

Produced from nndc.bnl.gov/ensdf

But the NNDC website doesn't do permalinks and cached results go away after a while, so if you wait you'll have to find it again.

Emilio Pisanty

7:27 PM

@rob works

rob

@EmilioPisanty Don't worry, it'll quit :-)

Emilio Pisanty

@rob not sure how to read it though

rob

First table is list of levels: energy, $J^P$, half-life. The $16^+$ state with the 31-year life is the last one.

Emilio Pisanty

@rob yeah, but how would that one decay?

rob

Second table is gamma rays, sorted by energy in keV, labeled by originating level and multipolarity (if known)

Emilio Pisanty

7:32 PM

ah

rob

So there's a very soft transition which is probably E3 and goes to the probably-$13^-$ state, cascading from there ...

Emilio Pisanty

So one E5 line by 587 down to the (11-) at 1859

heather

hello all

rob

@EmilioPisanty Yep

And an M4 to a (12^-) state at 1859

Emilio Pisanty

plus one M4/E5 line by 309 down to the (12-) at 2136

then

after which it'd presumably decay down much faster?

rob

7:34 PM

Parentheses indicate doubt about spin/parity assignments.

Emilio Pisanty

so the lowest-energy one would be the 309MeV one

rob

Yes, apart from the two isomers, all the other states probably decay more or less instantaneously.

Emilio Pisanty

@rob ah, so there's possibly one E3 line down to the level just under?

'cause that would drastically change the multipolarity argument

309MeV is about 1.6 fm

12.7 is about 8fm

rob

@EmilioPisanty 309 keV is about 1600 fm, you mean.

Qmechanic

@EmilioPisanty : Ok.

Emilio Pisanty

7:37 PM

no, wait

@rob all of these are in keV?

ah, yes

rob

@EmilioPisanty The referenced paper (2015KiAA: Phys Lett B750, 89 (2015)) may have information on the relative strengths of the different transitions --- I'm not sure why that would have been left out of the NNDC table.

Emilio Pisanty

309keV is about 640fm, 12.7keV is about 15,500fm

what size is a hafnium nucleus?

Is 1fm a reasonable figure?

rob

@EmilioPisanty You're off by 2π: 309 keV is 4000 fm

Emilio Pisanty

@rob hmmm, right you are

rob

@EmilioPisanty Nuclear radius is a few femtometers, but adjacent nuclei are separated by a few angstroms.

@EmilioPisanty As penance, we'll change the name of the chat room to "the h" for a week

Emilio Pisanty

7:43 PM

@rob but this is about the transition between states of the same nucleus, right? so the matrix elements are between states of the extent of a few fm

rob

@EmilioPisanty Are we back on lasers yet? If we want stimulated emission, we would want many isomers to decay in phase with each other.

Emilio Pisanty

@rob sure, but this is just the transition probability of the thing happening to begin with

right?

rob

All electromagnetic transitions in nuclei have wavelength longer than the diameter of the nucleus, because no nuclei are bigger than a few femtometers and no nuclei have more than a few MeV of nucleon separation energy.

Emilio Pisanty

@rob sure, but that doesn't mean that the nucleus isn't sensitive to the spatial derivatives of the mode

it just makes it less sensitive

by something like $a/\lambda$, where $a$ is the size of the system

rob

@EmilioPisanty The rule I remember is that there's a term in the transition probability like $(a/\lambda)^{\Delta J}$, where $a$ is nuclear size and $\lambda$ is transition wavelength.

Emilio Pisanty

7:48 PM

@rob yeah, precisely

rob

The E3 transition is so soft, at 12 keV, that it kind of makes sense for it to compete with the 20x more energetic E5 and M4 transitions.

Emilio Pisanty

for those transitions, I make $a/\lambda$ to be 2.5E-4 for the 309keV E5 transition, and 1E-5 for the 12keV (E3) transition

by that measure, the 12keV E3 transition would come out ahead by about 10^5

0celo7

What on earth is going on here

This sounds like physics

AccidentalFourierTransform

ugh, disgusting

Emilio Pisanty

because $(a/\lambda_{12\mathrm{keV}})^3\approx 10^{-15}$, and $(a/\lambda_{309\mathrm{keV}})^5\approx 10^{-18}$

@0celo7 pretty much

we're discussing metastable hafnium

just, you know, because.

Hafnium controversy

The hafnium controversy is a debate over the possibility of 'triggering' rapid energy releases, via gamma ray emission, from a nuclear isomer of hafnium, 178m2Hf. The energy release is potentially 5 orders of magnitude (105 times) more energetic than a chemical reaction, but 3 orders of magnitude less than a nuclear reaction. In 1998, a group led by Carl Collins of the University of Texas at Dallas reported having successfully initiated such a trigger. Signal-to-noise ratios were small in those first experiments, and to date no other group has been able to duplicate these results. Peter Zimmerman...

rob

7:55 PM

@EmilioPisanty Hmmm, I did different.

I used $a = (178)^{1/3}\rm fm = 6 fm$

For the 12 keV E3 transition I get $(a/λ)^3 = 2\times10^{-13}$

for the 590 keV E5 transition I get $(a/λ)^5 = 2\times10^{-13}$

Emilio Pisanty

yeah, I get the same

wait, so the 590 keV is more likely than the 309keV?

duh, yes, of course

rob

The 309 keV transition is a competition between M4 and E5. The magnetic transitions go like $(a/λ)^{\Delta J + 1}$, which is why the M4 and E5 compete. So I think that's the winner by a factor of some small power of two.

No, you're right --- the larger-energy one is the winner.

Emilio Pisanty

ok, but I'm sold on the fact that this is a pretty damn unlikely transition

rob

The 309 keV transition is in third place intensity-wise.

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