1) A particular smoke detector contains 1.35 ? Ci of 241Am, with a half-life of
ID: 1063485 • Letter: 1
Question
1)
A particular smoke detector contains 1.35 ?Ci of 241Am, with a half-life of 458 years. The isotope is encased in a thin aluminum container. Calculate the mass of 241Am in grams in the detector.
Express your answer numerically in grams.
2)
BNCT relies on the initial targeting of tumor cells by an appropriate chemical compound tagged with 10 5B, which preferentially concentrates in tumor cells. During the irradiation of the tumor site by neutrons (10n) the 10 5B absorbs a low-energy neutron (10n), and it ejects an energetic short-range alpha particle (42? or 42He) and lithium ion along with gamma radiation (?). This radiation deposits most of its energy within the cell containing the original 10 5B atom. Therefore, if a higher concentration of 10 5B exists in tumor cells relative to other normal tissues, a concomitantly higher dose will be delivered to the tumor cells during neutron irradiation.
What is the nuclear reaction that takes place in the tumor cell?
Explanation / Answer
1) The specific activity of a nuclide is given by
SA = ln(2) * NA / (T½ * u)
where
ln(2) is the natural logarithm of 2
NA = Avogadro's Number
T½ is the half-life
u is the atomic mass
For Am-241, we need to convert the half-life to years, so multiply by 31.56 megaseconds.
458*31.56e6 = 14.45e9 seconds
For Am-241, we can approximate u = 241, or look it up on Wikipedia and use 241.0547. (Note the approximation is off by 0.022%, so it's not really worth worrying about.)
SA = 0.6931 * 6.02e23 / (14.45e9 * 241)
= 4.1724e23 / 3.482e12
= 1.1983e11 sec^-1 or 1.1983e11 Bq.
One Curie is 3.7e10 Bq, so one gram of Am-241 is
1.1983e11 / 3.7e10 = 3.239 Ci/g
So 2.45e-6 µCi / 3.239 Ci/g = 0.7564µg
2)
10 5B + 1 0n ? 4 2He + 7 3Li + ?
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