Can I consider them simply short (less than second at least)? Is it the reason why some are unknown (no value given)? Also how can I convert MeV to
time? Using e=mc2 formula, correct? Why are they given in MeV?j_sum1 - 19-4-2016 at 14:19
You can convert between energy and mass. This is entirely valid in some contexts.
You can convert between time and distance -- also valid in some contexts, but the conversion is not symmetrical. Time has some properties that
distance does not -- namely that it appears to go in one direction only. Second law of thermodynamics and all that.
You cannot convert between energy and time. They have different units and even in relativity are quite different quantities.MrHomeScientist - 20-4-2016 at 10:22
I've never seen a half life given in MeV. Can you give an example link?unionised - 20-4-2016 at 12:40
But maybe when they say that half-life is independant of temperature, they only think about low temperatures. But what if I heat (accelerate) proton
(H1) to energy of 782.38 keV or 1804.38 keV. Will it become neutron? And fuse it with proton to get deuterium, or fuse it with anything else to get
any isotope? Isn't just one of these energies enough to make neutron stable (if we could hold it isolated to not react with anything)? I mean neutron
has half-life few minutes and has exactly same energy as above (just opposite sign) where it becomes proton (782.38 keV)! www.nndc.bnl.gov/chart/decaysearchdirect.jsp?nuc=1NN&unc... nrv.jinr.ru/nrv/webnrv/map/nucleus.php?q=H1
[Edited on 20-4-2016 by radioboy]Theoretic - 20-4-2016 at 23:43
But maybe when they say that half-life is independant of temperature, they only think about low temperatures. But what if I heat (accelerate) proton
(H1) to energy of 782.38 keV or 1804.38 keV. Will it become neutron? And fuse it with proton to get deuterium, or fuse it with anything else to get
any isotope?
When two protons fuse in the proton-proton reaction inside the sun, then the energy released from their attraction by the strong nuclear force, does
transform one of the protons into a neutron. And the result is a deuterium nucleus, as you say
And in decays of some nuclei, the proton does either capture an electron from one of the atoms electron shells, *or* decays and emits a positron, in
both cases turning into a neutron.
One can even drive this process purposely, and bombard nuclei with electrons of a specific energy; which both supplies the necessary energy, and the
electron to react with the proton, and create a neutron; thus creating a different isotope.
[Edited on 21-4-2016 by Theoretic]radioboy - 21-4-2016 at 03:05
I also realized that electron capture (not beta+) is opposite of beta- decay. Just can't figure half-life of any induced decay by energy (as proton to
neutron conversion, and does high temperature really affects it). Also, it seems much more simple and faster to simply fuse proton with something
because coulomb barrier is the lowest for hydrogen isotopes. Even lower than energy for making neutron from proton. And probably fast.
[Edited on 21-4-2016 by radioboy]Theoretic - 21-4-2016 at 04:35
I also realized that electron capture (not beta+) is opposite of beta- decay. Just can't figure half-life of any induced decay by energy (as proton to
neutron conversion, and does high temperature really affects it). Also, it seems much more simple and faster to simply fuse proton with something
because coulomb barrier is the lowest for hydrogen isotopes. Even lower than energy for making neutron from proton. And probably fast.
[Edited on 21-4-2016 by radioboy]
And there is *no*coulomb barrier for electrons.
When there's a half life is listed together with an energy, that's probably just the decay energyMrHomeScientist - 21-4-2016 at 05:36
Agree with Theoretic. It's certainly the decay energy. You can't convert between energy and time, it's nonsensical. PhDChem - 9-6-2016 at 00:29
Agree with Theoretic. It's certainly the decay energy. You can't convert between energy and time, it's nonsensical.
In CRC Handbook of Chemistry and Physics it says "Half-Life/Resonance Width: Half-life in decimal notation, microseconds, milliseconds, seconds,
minutes, hours, days, years. For quasi-stable nuclides, the measured width at half-maximum of the energy resonance is given."
And they gave half-life in MeV, keV, and eV for some nuclides (H4, H5, H6, H7...) for which I believe have extremely short half-lives (am I right?).
What means "resonance width" and "quasi-stable nuclide"?Polverone - 9-6-2016 at 12:48
The book Be Beryllium: The Element. Production, Atom, Molecules, Chemical Behavior, Toxicology says that
Beryllium 6 is particle unstable with respect to helium 4 + 2 protium by 1371 +- 5 keV. The width of the ground state for this decay is 92
+- 6 keV in the center-of-mass system. (My emphasis).
These half-lives reported in energy units are really reporting "the width of the ground state" for the decay? I don't know what that means with
respect to time units. But presumably there is some relationship to conventional time units since reputable data sources keep
reporting half-lives in these energy units.
The primary source for the beryllium 6 data is <a href="http://sci-hub.cc/10.1016/0375-9474(84)90650-X">Energy Levels of Light Nuclei
A=5-10</a>, Nuclear Physics A 413 1/214, by F. Ajzenberg-Selove.Marvin - 10-6-2016 at 01:42
[Edited on 10-6-2016 by Marvin]chornedsnorkack - 21-6-2016 at 03:07
You cannot convert between time and energy in classical relativity, but you can in quantum mechanics - even in nonrelativistic quantum mechanics.
Heisenberg uncertainty principle and Planck constant.
The practicably measurable lifetimes span over 50 orders of magnitude, from 10-23 seconds to 1030 seconds or so.
For short lifetimes, decay width is a useful measurement. It also is distinct from decay energy. For example, beryllium 8 undergoes alpha decay. Decay
energy 188 keV (94 per each alpha particle), devay width about 7 eV. Meaning halflife about 10-16 seconds.
