- #1
overpen57mm
- 7
- 0
- Homework Statement
- A 50.0000 g bone sample was placed in a nuclear reactor core such that it was
completely and homogeneously exposed to a neutron field. The purpose of placing
this bone sample in the core was to determine the weight fraction of total calcium in
the bone sample. The only reaction that occurs between the calcium in the bone
sample, for the neutron energy in this core, is the 48 Ca(n,γ) 49 Ca reaction. This
reaction is read as follows: A nucleus of 48 Ca reacts with one neutron, releases a γ-
ray and is converted to 49 Ca in the process. 48 Ca comprises 0.2 % of total calcium
in bone. Presume that the neutron fluence rate in the reactor core is 10 6 neutrons
cm −2 s −1 and that the cross section for this reaction is 2 barns. The sample was
irradiated for 4 minutes in the core, was removed from the core and was allowed to
sit for 5 minutes before the activity of
49 Ca was determined to be 166 kBq.
Determine the weight fraction of total calcium in this bone sample.
- Relevant Equations
- Specific Activity for calcium-49
Atomic mass for calcium 40 and 48
Exponential decay
Nuclear cross section formula
I think I have the answer, but when I calculate it out, I'm off by at least an order of magnitude.
My process goes like this:
The (looked up) specific activity of Ca-49 is ##1.63*10^7 TBq/g##, which is ##1.63*10^{16} KBq/g##
The activity of the sample after 5 minutes rest is ##166KBq##, per the question, so the mass of the Ca-49 in the sample at that time is
##A/A_sp = {166} / {1.63*10^{16}} = 1.018*10^{-14}g##
The decay constant (also looked up) of Ca-49 is ##0.001325##, so if it rested 5 minutes (300 seconds), then the initial mass would be
##N_1 / {e^{-λt}} = {1.018*10^{-14}g} / {e^{-(0.001325)(300)}} = 1.5149*10^{-14}g##
This is the mass of Ca-49 at the moment the sample is pulled from the reactor.
It might be important to note that I don't consider the decay of Ca-49 while inside the reactor. This might be an error but even if it is I don't think it will give me an answer as wrong as my current one.
Having the mass of Ca-49, I move onto the reaction taking place. The nuclear cross-section per the question is ##2 barn##, which is ##2*10^{-24}cm^2/nucleus##, the fluency of neutrons in the reactor is ##10^6 neutrons/{cm^2s}##.
Since ##σ=R/I##, ##σI=R##, ##{10^6 neutrons/{cm^2s}} * {2*10^{-24} cm^2/nucleus} = {2*10^{-18} reactions/{s nucleus}}##
The question doesn't give the nuclear cross section in terms of nucleus, but surely if I increase the number of nuclei, then the reaction speed must increase proportionally? This is how I understand nuclear cross-section, please correct me if I'm wrong.
Since the sample is in the reactor for 4 minutes (240 seconds), I multiply the reactions per second per nucleus to give: ##2*10^{-18} 1/{s nucleus} * 240s = 4.8*10^{-16} reactions/nucleus##.
Back to Ca-49 for a moment, the atomic mass of Ca-49 is ##48.995g/mol##, so ##1.5149g## equals ##1.861*10^8 atoms##
The number of Ca-49 atoms is the number of reactions that took place, since that's the only way they can form, so since I have the number of reactions per nuclei, and the number of reactions, I should be able to get the number of nuclei like so: ##1.861*10^8 reactions / {4.8*10^{-16} reactions/nucleus} = 3.877*10^23 nuclei##.
The problem becomes apparent here, that many calcium-48 nuclei is already 30 grams, and since according to the question only 0.2 percent of all calcium is Ca-48, the total mass of calcium would be well over the 50g of bone I started with, over 1.5kg. This can't be, so I must have gone wrong somewhere.
For what it's worth I went over this with my father, who's a PHD (in physics, but not nuclear physics), and he was stumped as well, so you are my only hope.
My process goes like this:
The (looked up) specific activity of Ca-49 is ##1.63*10^7 TBq/g##, which is ##1.63*10^{16} KBq/g##
The activity of the sample after 5 minutes rest is ##166KBq##, per the question, so the mass of the Ca-49 in the sample at that time is
##A/A_sp = {166} / {1.63*10^{16}} = 1.018*10^{-14}g##
The decay constant (also looked up) of Ca-49 is ##0.001325##, so if it rested 5 minutes (300 seconds), then the initial mass would be
##N_1 / {e^{-λt}} = {1.018*10^{-14}g} / {e^{-(0.001325)(300)}} = 1.5149*10^{-14}g##
This is the mass of Ca-49 at the moment the sample is pulled from the reactor.
It might be important to note that I don't consider the decay of Ca-49 while inside the reactor. This might be an error but even if it is I don't think it will give me an answer as wrong as my current one.
Having the mass of Ca-49, I move onto the reaction taking place. The nuclear cross-section per the question is ##2 barn##, which is ##2*10^{-24}cm^2/nucleus##, the fluency of neutrons in the reactor is ##10^6 neutrons/{cm^2s}##.
Since ##σ=R/I##, ##σI=R##, ##{10^6 neutrons/{cm^2s}} * {2*10^{-24} cm^2/nucleus} = {2*10^{-18} reactions/{s nucleus}}##
The question doesn't give the nuclear cross section in terms of nucleus, but surely if I increase the number of nuclei, then the reaction speed must increase proportionally? This is how I understand nuclear cross-section, please correct me if I'm wrong.
Since the sample is in the reactor for 4 minutes (240 seconds), I multiply the reactions per second per nucleus to give: ##2*10^{-18} 1/{s nucleus} * 240s = 4.8*10^{-16} reactions/nucleus##.
Back to Ca-49 for a moment, the atomic mass of Ca-49 is ##48.995g/mol##, so ##1.5149g## equals ##1.861*10^8 atoms##
The number of Ca-49 atoms is the number of reactions that took place, since that's the only way they can form, so since I have the number of reactions per nuclei, and the number of reactions, I should be able to get the number of nuclei like so: ##1.861*10^8 reactions / {4.8*10^{-16} reactions/nucleus} = 3.877*10^23 nuclei##.
The problem becomes apparent here, that many calcium-48 nuclei is already 30 grams, and since according to the question only 0.2 percent of all calcium is Ca-48, the total mass of calcium would be well over the 50g of bone I started with, over 1.5kg. This can't be, so I must have gone wrong somewhere.
For what it's worth I went over this with my father, who's a PHD (in physics, but not nuclear physics), and he was stumped as well, so you are my only hope.
Last edited:
