PASSAGE. The remaining multiple-choice questions refer to the following passage.
ID: 1479871 • Letter: P
Question
PASSAGE. The remaining multiple-choice questions refer to the following passage.
Different isotopes of Iodine (atomic number 53) are used in medicine. While low doses of the radioisotope Iodine-131 result in a detectable increase in thyroid cancer, high doses of the isotope paradoxically are sometimes less dangerous than low doses, since they tend to kill thyroid tissues which would otherwise become cancerous as a result of the radiation. For example, children treated with moderate dose of I-131 for thyroid adenomas had a detectable increase in thyroid cancer, but children treated with a much higher dose did not. Thus, iodine-131 is increasingly less employed in small doses in medical use (especially in children), but it is still successfully used in large and maximal treatment doses, as a way of killing targeted tissues. One such therapeutic use of high doses of I-131 is employed as a treatment for Graves disease.
Iodine-131 radioisotopes are produced from neutron bombardment of Tellurium-130 in a nuclear reactor followed by a radioactive decay to I-131:
Te-130 + n Te-131
Te-131 I-131 + e-
The iodine-131 radioisotope decays in a beta decay with a half-life of 8.0 days.
1.
What is the number of protons and neutrons in Te-131?
A) 53 protons and 131 neutrons
B) 52 protons and 131 neutrons
C) 53 protons and 78 neutrons
D) 54 protons and 77 neutrons
E) 52 protons and 79 neutrons
2.
The decay of Te-131 is ____
A) a positron decay
B) an alpha decay
C) a beta decay
D) an electron capture
E) a gamma decay
3.
A patient is injected with I-131 with a decay rate of R. How many days later will the decay have dropped to 5% of R?
A) 8.0 days
B) 160 days
C) 34.5 days
D) 40.0 days
E) 32.0 days
4.
Which isotope is produced in the beta decay of I-131?
A) Xe-130
B) Xe-131
C) Te-130
D) Te-131
E) Xe-132
Explanation / Answer
1. Atomic number of Te = 52
Atomic mass = 131
So number of proton = 52
Number of neutron = 131 - 52 = 79
Answer. E) 52 protons and 79 neutrons
2. C) a beta decay (an electron is ejected)
3. Half life = 8 days
decay at time t = decay at time 0 * 2 ^ (-t / half time)
=> 0.05 = 2^(-t/8)
=> t = 34.58 days
Answer. C) 34.5 days
4. Beta decay of I-131 gives Xe-131
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