Iodine in the body is preferentially taken up by the thyroid gland. Therefore, radioactive iodine in small doses is used to image the thyroid and in large doses is used to kill thyroid cells to treat some types of cancer or thyroid disease. The iodine isotopes used have relatively short half-lives, so they must be produced in a nuclear reactor or accelerator. One isotope frequently used for imaging is I123; it has a half-life of 13.2 h and emits a 0.16-MeV gamma-ray photon. One method of producing I123is in the nuclear reaction T123e+pI123+n. The atomic masses relevant to this reaction are T123e, 122.904270 u; I123, 122.905589 u; n, 1.008665 u; and H1, 1.007825 u. The iodine isotope commonly used for treatment of disease is I131, which is produced by irradiating T130ein a nuclear reactor to form T131e. The T131ethen decays to I131. I131undergoes beta- decay with a half-life of 8.04 d, emitting electrons with energies up to 0.61 MeV and gamma-ray photons of energy 0.36 MeV. A typical thyroid cancer treatment might involve administration of 3.7 GBq of I131.

Which reaction produces T131e in the nuclear reactor?

(a) T130e+nT131e;

(b) I130+nT131e;

(c) T132e+nT131e;

(d) I132+nT131e.

Short Answer

Expert verified

The reaction produces T131eis:localid="1664202652275" T52130e+nT52131e.

Step by step solution

01

Conservation of atomic number

First, we write the reaction equation with the unknown particle XZAfollows

T52130e+XZAT52131e

From the conservation of atomic number and the number of nucleons, we get

52+Z52Z=0130+A131A=1

The particle with Z = 0 and A = 1 is the neutron.

02

Conclusion

Therefore, the answer is : T52130e+nT52131e.

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