Dear Meghnath,
> On 02 May 2016, at 07:21, meghms_at_barc.gov.in wrote:
> Actually I have measured the activity from the peak corresponding to the
> 411 KeV of gamma photon using HPGe detector.
In this case you have to consider only the channels giving contribution to the 411 keV peak.
> According to your suggestion i.e. "Residual nuclei are always considered
> at the ground state while the isomeric excited states of gold, with their
> activity toward the decay to the ground state".
> From the last line of your statement what I understand is that this
> isomer(198Au) is surely emitting photons.
> Now, if 197Au absorbs a neutron then other than decaying through beta
> followed by gamma is there any other way of decay to come to ground
> state!!!
Now, I’m not really in your field but to my knowledge when a ^{197}_{79}Au does not absorb the neutron and still gets excited (first level ) become a ^{197m}_{79}Au (nuclear isomer). Similar if ^{197}_{79}Au absorb a neutron and the result nucleus is created in an excited (first) level it will become ^{198m}_{79}Au.
Photons from those transitions are typically lower that 1keV and I would say can be neglected for your simulation.
>
> Again, I did the same simulation in vacuum for the pure thermal neutron
> (0.025 eV)where the I got only Residual nuclei and no ISOMER (198Au)was
> formed. Is it forming only for high energy neutrons!!!
Sure because it’s an excited nuclear isomer state…
Cross sections are smalls, for ^{197m}_{79}Au you have a 100% error, and for ^{198m}_{79}Au around 8%!
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Received on Mon May 02 2016 - 15:12:38 CEST