# Neutron activation question

1. Jun 21, 2009

### cortiver

1. The problem statement, all variables and given/known data\
(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).

2. Relevant 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
$$\frac{dN^{*}}{dt} = -\lambda N^{*}$$.

3. 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?

Last edited: Jun 21, 2009
2. Jun 21, 2009

### Astronuc

Staff Emeritus
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?

3. Jun 21, 2009