Nuetron activation


by ineedhlp
Tags: activation, neutron activation, neutron flux, nuetron
ineedhlp
ineedhlp is offline
#1
Nov27-13, 04:57 PM
P: 3
Co-60 is produced from Co-50 through neutron activation. Neutron flux 10^18 nuetrons/(s*m^2). cross section is 20 barns and the mass of Co-50 40mg. How many Co-60 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 CO-50 which was 4.08598x10^20 atoms. Then i multiplied the amount of atoms by the number of neutrons in Co-59 and add to the number of neutrons found by the flux and then divided by 33 neutrons for how many are in C0-60. This gave me the number of nuclei of Co-60 found which was not the right answer.
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mfb
mfb is offline
#2
Nov27-13, 05:28 PM
Mentor
P: 10,809
I think C-50 should be C-59 everywhere.

I multiplied the flux by the seconds in one week and by 20 barns in m^2 and got 0.0122 neutrons.
Careful, the unit is more like "neutrons per atom".

Then i multiplied the amount of atoms by the number of neutrons in Co-59
Why?
The reaction involves the whole nucleus, and the cross-section is "per nucleus".

and then divided by 33 neutrons for how many are in C0-60.
Again, why?
ineedhlp
ineedhlp is offline
#3
Nov27-13, 05:53 PM
P: 3
I assumed that if i had the total number of neutrons of the CO-59 and added it to the number of neutrons that came from the flux, i could find the total number of Co-60 neutrons and then find how many nuclei from that.

mfb
mfb is offline
#4
Nov27-13, 06:12 PM
Mentor
P: 10,809

Nuetron activation


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


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