A) Neutron activation analysis for a sample of a rock revealed the presence of 5

Question

A) Neutron activation analysis for a sample of a rock revealed the presence of 5926Fe ,which has a half-life of 46.3 days . Assuming the isotope was freshly separated from its decay products, what is the mass of 5926Fe in a sample emitting 1.00 mCi of radiation?

B) 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

A) decay constant = ln2/half life = 0.693/46.3 = 0.015 day-1 = 1296 s-1

Now, 1 mCi = 3.7*107 decays per second

**** data is mi ssing

B) 10B5 + 1n0 ------> 7Li3 + 4He2 +

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