How Do You Calculate Neutron Flux in Neutron Activation Analysis?

AI Thread Summary
The discussion focuses on calculating neutron flux in neutron activation analysis using a past exam question involving Na-24 production from Al-27 irradiation. Participants highlight the need for specific values, such as the half-life of Na-24 and the atomic mass of aluminum, which were not provided in the exam. The decay constant (λ) can be derived from activity (A) using the relationship A = λN, but without knowing the neutron flux (Φ) or the number of Al atoms (N), calculations are challenging. The conversation emphasizes the importance of having all necessary data for accurate calculations in nuclear physics problems. Ultimately, the lack of complete information complicates the solution process.
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Homework Statement

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(From a past exam paper)
Na-24 can be produced by irradiation of 27-Al with energetic neutrons. It decays by beta emission.

An aluminium sample of mass m=0.2g was irradiated by energetic neutrons for a period of 1 hour. Two hours after the end of irradiation the total 24-Na activity of the sample was 295 Bq.
(i) Calculate the total number of Na-24 nuclei that were present in the aluminium sample immediately at the end of the irradiation.
(ii) What was the neutron flux (use a reaction cross-section of \sigma=0.125 barns).


Homework Equations


Basically, when the sample is being irradiated the number of 24-Na atoms N^{*} obeys
\frac{dN^{*}}{dt} = R - \lambda N^{*}
where R = N\sigma \Phi is the rate of neutron absorption, N is the number of atoms in the sample, \sigma is the microscopic cross-section, and \Phi is the neutron flux. \lambda is the decay constant. Once the irradiation is stopped, it decays according to
\frac{dN^{*}}{dt} = -\lambda N^{*}.

The Attempt at a Solution


It seems to be just a simple plug-and-chug question based on the rudimentary treatment of neutron activation analysis that we did in class, but the thing is that I can't see how you can possibly do it without knowing the Na-24 half-life and the atomic mass of aluminium, and neither of these values were given anywhere in the exam paper. Am I missing something obvious here?
 
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The atomic mass of Al-27 is 27 amu. To go from 27Al to 22Na requires an (n,α) reaction.

The half-life of Na-22 is 14.951 hrs. One reference is http://www.nndc.bnl.gov/chart/.

However, since A = λ N, then is it possible to determine λ, from the information given?
 
Thank you for your answer.
Astronuc said:
The atomic mass of Al-27 is 27 amu. To go from 27Al to 22Na requires an (n,α) reaction.
You're right, I didn't realize that the mass defects are so small that you can just ignore them for these purposes.
The half-life of Na-22 is 14.951 hrs. One reference is http://www.nndc.bnl.gov/chart/.
But the point is, this question was in a past exam paper. That reference would not have been available.
However, since A = λ N, then is it possible to determine λ, from the information given?
But we don't know N either, since the neutron flux is unknown.
 
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