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Nuetron activation 
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#1
Nov2713, 04:57 PM

P: 3

Co60 is produced from Co50 through neutron activation. Neutron flux 10^18 nuetrons/(s*m^2). cross section is 20 barns and the mass of Co50 40mg. How many Co60 are produced in one week?
My attempt: I multiplied the flux by the seconds in one week and by 20 barns in m^2 and got 0.0122 neutrons. Then found how many atoms are in the 40mg of CO50 which was 4.08598x10^20 atoms. Then i multiplied the amount of atoms by the number of neutrons in Co59 and add to the number of neutrons found by the flux and then divided by 33 neutrons for how many are in C060. This gave me the number of nuclei of Co60 found which was not the right answer. 


#2
Nov2713, 05:28 PM

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P: 11,917

I think C50 should be C59 everywhere.
The reaction involves the whole nucleus, and the crosssection is "per nucleus". 


#3
Nov2713, 05:53 PM

P: 3

I assumed that if i had the total number of neutrons of the CO59 and added it to the number of neutrons that came from the flux, i could find the total number of Co60 neutrons and then find how many nuclei from that.



#4
Nov2713, 06:12 PM

Mentor
P: 11,917

Nuetron activation
You don't have the total number of neutrons in the beam, and you don't need it.



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