Difference between revisions of "Ni-08-22-13"

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Since the two neutron knockout reaction requires a higher photon energy one may be able to estimate the bremmsstrahlung distribution by requiring that the ratio of the integral for photon energies allowing single neutron knockout to the integral for two neutron knockout match the observed counting rate that has included cross -section
 
Since the two neutron knockout reaction requires a higher photon energy one may be able to estimate the bremmsstrahlung distribution by requiring that the ratio of the integral for photon energies allowing single neutron knockout to the integral for two neutron knockout match the observed counting rate that has included cross -section
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 +
Energy needed to liberate a neutron from the Nucleus
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Neutron Separation Energy=<math>S_n = B({A \atop Z} X_{N})-B({{A-1} \atop Z} X_{N-1})</math>
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Revision as of 21:40, 16 November 2014

Nickel foils were placed in front of and behind the Au+Sand, the blank crucible, and the gold foil. Nickel foils were also place in front of and behind the crucible with the Barium Oxide.

lines are observed near, 238, 511, 578, 604, 903, 1264, 1362 keV

Naturally occurring Isotopes of Nickel are

Ni-58 ( 68.077% ) Ni-60 ( 26.223% ) Ni-61 ( 1.140% ) Ni-62 ( 3.634% ) Ni-64 ( 0.926% )


Neutron knockout

Ni-58 ->Ni-57 has the following lines in order of decreasing intensity: 511, 1377, 127 , 1919 half life = 35.6 hours

Ni-60 -> Ni-59 : no high intensity lines above 100 keV, half life 7e4 years

Proton knockout

Ni-58-> Co-57 : 122,136, half life 271.74 days

N-P knockout

Ni-58->Co-56 :847, 1238, 511, 2599, 1771, 1038, half life = 77 days


Extracting Photon flux using Nickel Foil

The two reactions below can be used to extrapolate the photon flux


[math]{58 \atop 28 }Ni_{76}(\gamma,n){57 \atop 28 }Ni_{75}[/math]
[math]{58 \atop 28 }Ni_{76}(\gamma,2n){56 \atop 28 }Ni_{74}[/math]


Since the two neutron knockout reaction requires a higher photon energy one may be able to estimate the bremmsstrahlung distribution by requiring that the ratio of the integral for photon energies allowing single neutron knockout to the integral for two neutron knockout match the observed counting rate that has included cross -section

Energy needed to liberate a neutron from the Nucleus

Neutron Separation Energy=[math]S_n = B({A \atop Z} X_{N})-B({{A-1} \atop Z} X_{N-1})[/math]


PAA_8-22-13#Nickel_normalization_foils

A_W_thesis