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Chapter 25: Problem 44

Iodine-131 \(\left(t_{1 / 2}=8.04 \text { days }\right),\) a \(\beta\) emitter, is used to treat thyroid cancer. (a) Write an equation for the decomposition of \(^{131}\) I. (b) If you ingest a sample of NaI containing \(^{131}\) I, how much time is required for the activity to decrease to \(35.0 \%\) of its original value?

Short Answer

Expert verified
(a) Decay equation: ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- + \overline{\nu}_e . (b) Time for 35% activity: 13.9 days.

Step by step solution

01

Write the Nuclear Decay Equation

Iodine-131 ( ^{131}I ) undergoes beta decay to form Xenon-131 ( ^{131}Xe ). During beta decay, a neutron in the iodine nucleus is transformed into a proton, emitting a beta particle ( \beta^- ) and an antineutrino ( \overline{u}_e ). The nuclear equation for this decay is:\[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- + \overline{u}_e\]
02

Understand the Half-Life Formula

The half-life ( t_{1/2} ) of a radioactive isotope is the time it takes for half of the isotope to decay. For iodine-131, t_{1/2} = 8.04 days. The general formula to calculate the remaining activity after a certain time is:\[ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\]where A is the remaining activity, A_0 is the initial activity, t is the time elapsed, and t_{1/2} is the half-life.
03

Calculate the Time for 35% Activity

Since we want the activity to be 35.0\% of the initial activity (A/A_0 = 0.35) , substitute this into the formula and solve for t :\[ 0.35 = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\]Taking the natural logarithm on both sides, we get:\[ \ln(0.35) = \frac{t}{t_{1/2}} \cdot \ln\left(\frac{1}{2}\right)\]Rearranging for t , we find:\[ t = \frac{\ln(0.35)}{\ln(0.5)} \cdot t_{1/2}\]Substitute t_{1/2} = 8.04 days, we get:\[ t \approx \frac{\ln(0.35)}{\ln(0.5)} \cdot 8.04\]Calculate the numeric value to find t \approx 13.9 days.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iodine-131
Iodine-131 is a radioactive isotope commonly used in medical treatments, especially for thyroid cancer. Its primary role is to destroy cancerous cells in the thyroid gland due to its radioactive nature. This element has an atomic number of 53, indicating the presence of 53 protons in its nucleus.

Iodine-131 is notable for its relatively short half-life of 8.04 days, which makes it suitable for medical applications since it doesn't remain radioactive in the body for an extended period. Its rapid decay helps minimize long-term radiation exposure to the patient.
Beta Decay
Beta decay is a type of radioactive decay involving the transformation of a neutron into a proton within the nucleus. As a result, the atom emits a beta particle, which is high-energy and high-speed electron or positron.

When iodine-131 undergoes beta decay, it becomes xenon-131. This process increases the atomic number by one, transforming iodine with atomic number 53 into xenon with atomic number 54. Alongside the beta particle (denoted as \(\beta^-\)), a neutral subatomic particle called an antineutrino (\(\overline{u}_e\)) is also emitted. This decay helps us understand how iodine-131 turns into xenon-131.
Half-life Calculation
The half-life of a radioactive isotope is crucial in determining how long a substance remains active. For iodine-131, its half-life is 8.04 days, which means the amount of iodine-131 present will reduce by half every 8.04 days.

The formula for calculating how much of the isotope remains after a given time is:
  • \( A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \)
Where:
  • \(A\) is the remaining activity.
  • \(A_0\) is the initial activity.
  • \(t\) is time elapsed.
  • \(t_{1/2}\) is the half-life.
This formula helps predict how much of a radioactive sample remains after any given period.
Radioactive Isotopes
Radioactive isotopes are atoms with an unstable nucleus that release radiation in the form of particles or electromagnetic waves until they transform into a more stable form.

These isotopes are essential in various fields, from medicine to archaeology. In medical treatments, like the use of iodine-131, they offer a focused way to target and destroy unhealthy cells. This is because they emit radiation that the body can absorb, directing energy at specific tissues to provide therapeutic effects. Each isotope, like iodine-131, has unique properties making it suitable for different applications depending on its half-life and type of decay.
Nuclear Equation
A nuclear equation mathematically represents a radioactive decay process. It shows the transformation of the original isotope and the products formed as a result of the decay.

For iodine-131 undergoing beta decay, the nuclear equation is:
  • \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- + \overline{u}_e \]
This equation conveys that:
  • The starting nucleus of iodine-131 \((^{131}_{53}I)\) decays.
  • Xenon-131 \((^{131}_{54}Xe)\) is formed as the product nucleus.
  • A beta particle \((\beta^-)\) and an antineutrino \((\overline{u}_e)\) are emitted as by-products.
Understanding nuclear equations is essential for comprehending how isotopes transform and the types of radiation they emit.

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Most popular questions from this chapter

A sample of wood from a Thracian chariot found in an excavation in Bulgaria has a \(^{14} \mathrm{C}\) activity of \(11.2 \mathrm{dpm} / \mathrm{g} .\) Estimate the age of the chariot and the year it was made. \(\left(t_{1 / 2} \text { for }^{14} \mathrm{C} \text { is } 5.73 \times 10^{3}\right.\)years, andthe activity of \(^{14} \mathrm{C}\) in living material is \(14.0 \mathrm{dpm} / \mathrm{g} .\) )

Deuterium nuclei \(\left(_{1}^{2} \mathrm{H}\right)\) are particularly effective as bombarding particles to carry out nuclear reactions. ng equations: (a) \(^{114}_{48} \mathrm{Cd}+\\_{2}^{1} \mathrm{H} \longrightarrow ?+\\_{1}^{1} \mathrm{H}\) (b) \(_{3}^{6} \mathrm{Li}+\\_{2}^{1} \mathrm{H} \longrightarrow ?+\\_{1}^{0} \mathrm{n}\) (c) \(\\_{40}^{20} \mathrm{Ca}+\\_{2}^{1} \mathrm{H} \longrightarrow_{11}^{38} \mathrm{K}+?\) (d) \(?+\\_{2}^{1} \mathrm{H} \longrightarrow_{30}^{65} \mathrm{Zn}+\gamma\)

Calculate the binding energy per mole of nucleons for iron-56. Masses needed for this calculation (in g/mol) are \(_{1}^{1} \mathrm{H}=1.00783,_{0}^{1} \mathrm{n}=1.00867,\) and \(\frac{56}{26} \mathrm{Fe}=55.9349 .\) Compare the result of your calculation to the value for iron-56 in the graph in Figure 25.4

Iodine-131 is used to treat thyroid cancer. (a) The isotope decays by \(\beta\) -particle emission. Write a balanced equation for this process. (b) Iodine-131 has a half-life of 8.04 days. If you begin with \(2.4 \mu \mathrm{g}\) of radioactive \(^{131} \mathrm{I},\) what mass remains after 40.2 days?

Tritium, \(_{1}^{3} \mathrm{H},\) is one of the nuclei used in fusion reactions. This isotope is radioactive, with a halflife of 12.3 years. Like carbon- \(14,\) tritium is formed in the upper atmosphere from cosmic radiation, and it is found in trace amounts on Earth. To obtain the amounts required for a fusion reaction, however, it must be made via a nuclear reaction. The reaction of \(_{3}^{6}\) Li with a neutron produces tritium and an \(\alpha\) particle. Write an equation for this nuclear reaction.
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