5 ms·

From the paper, the light is UV-C at around 140nm or 8.4 eV. But it has to be very precisely the right energy to cause the transition, since nuclear states don’

by tlb 2y ago

From the paper, the light is UV-C at around 140nm or 8.4 eV. But it has to be very precisely the right energy to cause the transition, since nuclear states don’t have any place to dump excess energy to.

jhart99 2y ago
Ahhh thank you! I was wondering why the energy had to be so precise. That makes a ton of sense why it has to be so accurate. What makes this transition so low energy? The only other atomic excited state I have any knowledge of is the iron excited state used in Mossbauer spectroscopy. That transition is much higher energy. Also that one has some coupling to the electronic state of the nucleus. Does this Thorium transition have some special reason that it isn't coupled to the electronic state?
- adrian_b 2y ago
  The radiation emitted when nuclei transition between their internal energy levels is known as gamma rays. The gamma rays normally have energies per photon many orders of magnitude greater than for visible light and also much greater than for X-rays (which are produced by electrons accelerated by very high voltages when hitting a target). The thorium 229 nucleus is the only one that can emit gamma rays that are so low in energy that their energy is not only lower than for X-rays, but it is also lower than for many sources of ultraviolet light. For instance the ultraviolet light used in state-of-the-art lithography for semiconductor manufacturing has much higher frequency (shorter wavelength), by about ten times. These gamma rays of the Th229 have a wavelength that is not much shorter than the 184-nm ultraviolet light that can be obtained with a mercury-vapor lamp. What is important is that for such a frequency/wavelength it is possible to build laser sources, which enables the design of an atomic clock that will use thorium 229 nuclei instead of neutral atoms or ions of other elements (like ytterbium, lutetium, strontium, aluminum).
  - bawana 2y ago
    Arent gamma, x, and uv ALL em radiation? What makes a gamma with a wavelength near uv still allow it to be called gamma? Why dont we say the nucleon emits uv at 148 when it transitions to its ground state?
    - snarkconjecture 2y ago
      You could say that as well. Gamma is often (usually?) defined as any radiation that originates from nuclear state transitions rather than electrons; this tends to be very high energy but can overlap the range of EM radiation from electronic transitions. Th229 is the extreme outlier.
- twic 2y ago
  I found a paper which measured the energy of the transition [0], but it doesn't talk about why it's so low. Might be a starting point if you have more time to read than i do, though! EDIT Hmm [1]: > Interestingly, the existence of a nuclear excited state of such low energy seems to be a coincidence and there is currently no conclusive theoretical calculation that allows to predict nuclear levels to this precision. And there is a paper with a ton of detail and some nice diagrams of energy levels [2], but i'm not sure it really gets at "why". [0] https://arxiv.org/abs/1905.06308 https://arxiv.org/abs/1905.06308 [1] https://link.springer.com/article/10.1140/epja/s10050-020-00263-0 https://link.springer.com/article/10.1140/epja/s10050-020-00... [2] https://iopscience.iop.org/article/10.1088/1361-6455/ab29b8 https://iopscience.iop.org/article/10.1088/1361-6455/ab29b8
pfdietz 2y ago
The Q of nuclear transitions is just insane (as reflected by their long half life, something in excess of 1700 seconds here for free atoms.) The uncertainty relationship is normally written as delta-p delta-x > hbar/2, but it can also be written as delta-t delta-E > hbar/2. So, if the half life is very long, delta-E can be very small. This fact is used in Mössbauer spectroscopy (recoilless gamma emission in solids). The peak is so sharp that it was famously used by Pound and Rebka to detect the gravitational red shift in the lab at Harvard in 1960, reaching 1% accuracy by 1964. https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment
andrewflnr 2y ago
Where do electron transitions usually dump excess energy?
- ChrisClark 2y ago
  Don't they usually create a photon with that energy?
  - smaddox 2y ago
    For free atoms, yes. For atoms in a crystal lattice (or other solid), it's quite common for electrons to decay through phonon interactions, i.e. by emitting vibrations (i.e. heat) to the lattice.
- infogulch 2y ago
  In general it's possible for electrons to jump to sub-orbitals which gives them a wider band of wavelengths that they can emit and absorb photons. The jumps between sub-orbitals are usually in microwave or radio bands.
moconnor 2y ago
Interesting... there must be some error tolerance though, right? So there can be _some_ excess energy - where do they dump that to and what is the tolerance?

by tlb 2y ago

jhart99 2y ago

Ahhh thank you! I was wondering why the energy had to be so precise. That makes a ton of sense why it has to be so accurate. What makes this transition so low energy? The only other atomic excited state I have any knowledge of is the iron excited state used in Mossbauer spectroscopy. That transition is much higher energy. Also that one has some coupling to the electronic state of the nucleus. Does this Thorium transition have some special reason that it isn't coupled to the electronic state?

adrian_b 2y ago
The radiation emitted when nuclei transition between their internal energy levels is known as gamma rays. The gamma rays normally have energies per photon many orders of magnitude greater than for visible light and also much greater than for X-rays (which are produced by electrons accelerated by very high voltages when hitting a target). The thorium 229 nucleus is the only one that can emit gamma rays that are so low in energy that their energy is not only lower than for X-rays, but it is also lower than for many sources of ultraviolet light. For instance the ultraviolet light used in state-of-the-art lithography for semiconductor manufacturing has much higher frequency (shorter wavelength), by about ten times. These gamma rays of the Th229 have a wavelength that is not much shorter than the 184-nm ultraviolet light that can be obtained with a mercury-vapor lamp. What is important is that for such a frequency/wavelength it is possible to build laser sources, which enables the design of an atomic clock that will use thorium 229 nuclei instead of neutral atoms or ions of other elements (like ytterbium, lutetium, strontium, aluminum).
- bawana 2y ago
  Arent gamma, x, and uv ALL em radiation? What makes a gamma with a wavelength near uv still allow it to be called gamma? Why dont we say the nucleon emits uv at 148 when it transitions to its ground state?
  - snarkconjecture 2y ago
    You could say that as well. Gamma is often (usually?) defined as any radiation that originates from nuclear state transitions rather than electrons; this tends to be very high energy but can overlap the range of EM radiation from electronic transitions. Th229 is the extreme outlier.
twic 2y ago
I found a paper which measured the energy of the transition [0], but it doesn't talk about why it's so low. Might be a starting point if you have more time to read than i do, though! EDIT Hmm [1]: > Interestingly, the existence of a nuclear excited state of such low energy seems to be a coincidence and there is currently no conclusive theoretical calculation that allows to predict nuclear levels to this precision. And there is a paper with a ton of detail and some nice diagrams of energy levels [2], but i'm not sure it really gets at "why". [0] https://arxiv.org/abs/1905.06308 https://arxiv.org/abs/1905.06308 [1] https://link.springer.com/article/10.1140/epja/s10050-020-00263-0 https://link.springer.com/article/10.1140/epja/s10050-020-00... [2] https://iopscience.iop.org/article/10.1088/1361-6455/ab29b8 https://iopscience.iop.org/article/10.1088/1361-6455/ab29b8

pfdietz 2y ago

The Q of nuclear transitions is just insane (as reflected by their long half life, something in excess of 1700 seconds here for free atoms.) The uncertainty relationship is normally written as delta-p delta-x > hbar/2, but it can also be written as delta-t delta-E > hbar/2. So, if the half life is very long, delta-E can be very small. This fact is used in Mössbauer spectroscopy (recoilless gamma emission in solids). The peak is so sharp that it was famously used by Pound and Rebka to detect the gravitational red shift in the lab at Harvard in 1960, reaching 1% accuracy by 1964. https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment https://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment

andrewflnr 2y ago

Where do electron transitions usually dump excess energy?

ChrisClark 2y ago
Don't they usually create a photon with that energy?
- smaddox 2y ago
  For free atoms, yes. For atoms in a crystal lattice (or other solid), it's quite common for electrons to decay through phonon interactions, i.e. by emitting vibrations (i.e. heat) to the lattice.
infogulch 2y ago
In general it's possible for electrons to jump to sub-orbitals which gives them a wider band of wavelengths that they can emit and absorb photons. The jumps between sub-orbitals are usually in microwave or radio bands.

moconnor 2y ago

Interesting... there must be some error tolerance though, right? So there can be _some_ excess energy - where do they dump that to and what is the tolerance?