Solve Neutron Flux Problem with 27Al(p, 3pn)24Na Reaction

[coregis]

Hello. I'm a little unsure of how to proceed on this problem... Here it is:

A 3-Gev proton flux is monitored by measuring 24Na activity induced in 25 microm (6.85 microg/cm2) aluminum foil via 27Al(p, 3pn)24Na reaction (for 3 Gev protons, cross section = 9.1 mb). Exactly 15 hr after the end of 15-minute irradiation, the activity in the aluminum monitor is measured with a calibrated end-window counter (efficiency= 0.037 in the geometrical arrangement used). The net counting rate is 27,430 counts/min. What is the average proton flux through the sample during the irradiation?

So what I did first was correct for efficiency and found the actual counts to be 741351 cpm or 12355 cps.

Then since the count was made 15 hrs after bombardment, the activity at the end of the bombardment was 24710 cps.

At this point, I'm stuck because I don't know how to handle the areal density and convert it into the number of atoms in the beam. Can this problem even be solved without knowing the area of the beam? I pretty sure I need to find the number of atoms of Co that are in the beam, right? Any help is appreciated. Thanks.

 

[wais]

Hi there,

Thank you for sharing your progress on this problem. It seems like you have made some good initial steps in solving it. I understand your confusion about the areal density and the number of atoms in the beam. Let me try to provide some guidance on how to proceed.

Firstly, you are correct that the areal density is needed in order to calculate the number of atoms in the beam. The areal density is a measure of the number of atoms in a unit area, and it is usually expressed in atoms/cm2. In this case, the areal density of the aluminum foil is given as 6.85 microg/cm2. We can use this value to calculate the number of aluminum atoms in the foil.

To do this, we can use the atomic mass of aluminum (27 g/mol) to convert the areal density from micrograms to grams. This gives us a total mass of 0.00000685 g of aluminum in the foil. Next, we can use the atomic weight of aluminum (27 g/mol) to calculate the number of moles of aluminum in the foil. This gives us a value of 2.54 x 10^-7 moles of aluminum.

Finally, we can use Avogadro's number (6.022 x 10^23 atoms/mol) to convert the number of moles to the number of atoms. This gives us a total of 1.53 x 10^17 aluminum atoms in the foil.

Now that we have the number of aluminum atoms in the foil, we can use the cross section of the reaction (9.1 mb) to calculate the number of 24Na atoms produced. The cross section is a measure of the likelihood of a reaction occurring, and it is usually expressed in millibarns (mb). To calculate the number of 24Na atoms, we can use the following equation:

Number of 24Na atoms = cross section x number of aluminum atoms

Substituting the values we have calculated, we get:

Number of 24Na atoms = 9.1 mb x 1.53 x 10^17 atoms = 1.39 x 10^18 24Na atoms

Now that we have the number of 24Na atoms produced, we can use the decay equation to calculate the initial activity at the end of the bombardment. The decay equation is given by:

Activity = initial activity x e^(-lambda x