Question

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 (42α or 42He) and lithium ion along with gamma radiation (γ). 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?

Answer 1

Answer:

$^{10}B + ^{1}n_{th} \rightarrow ^{4}He + ^{7}Li + 2.79MeV$

Explanation:

BNCT therapy consists in the irradiation of the atoms of ¹⁰B that are contained in the tissue that has the tumor cells, with thermal neutrons to produce alpha particles and ⁷Li nucleus, according to the following nuclear reaction:

$^{10}_{5}B + ^{1}_{0}n_{th} \rightarrow ^{4}_{2}He + ^{7}_{3}Li + 2.79MeV$  

Since the tissues also contain H and N atoms, the neutron irradiation produces secondary reactions in which proton particles and gamma radiation are produced:

$^{1}H + n_{th} \rightarrow ^{2}H + \gamma$

$^{14}N + n_{th} \rightarrow ^{14}C + p$

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