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Iodine- 131 is used in the diagnosis and treatment of thyroid disease and has a half-life of 8.0 days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing \(10 . \mu \mathrm{g}^{131 \mathrm{I}}\) how long will it take for the amount of \(^{131} \mathrm{I}\) to decrease to 1\(/ 100\) of the original amount?

Short Answer

Expert verified
It will take approximately 53.26 days for the amount of Iodine-131 to decrease to 1/100th of the original amount.

Step by step solution

01

Identify the given values from the problem

We are given the following data: - \(N_0\) (initial amount consumed) = \(10\mu g\) - \(T_{1/2}\) (half-life of Iodine 131) = 8.0 days - We need to find the time \(t\) for the amount \(N(t)\) to decrease to 1/100th of the original amount, which is \(N(t) = \frac{N_0}{100}\).
02

Substitute the values into the exponential decay formula

We know that \(N(t) = N_0 \cdot (1/2)^{\frac{t}{T_{1/2}}}\), and we want to find the time \(t\) when \(N(t) = \frac{N_0}{100}\). Plugging in the values, we get: \[\frac{N_0}{100} = N_0 \cdot (1/2)^{\frac{t}{8}}\]
03

Simplify and solve for t

To solve for \(t\), we'll proceed as follows: 1. Divide both sides by \(N_0\): \[\frac{1}{100} = (1/2)^{\frac{t}{8}}\] 2. Take the natural logarithm of both sides to get the exponent down: \[\ln \left(\frac{1}{100}\right) = \ln \left( (1/2)^{\frac{t}{8}}\right)\] 3. Use the logarithm property to bring down the exponent: \[\ln \left(\frac{1}{100}\right) = \frac{t}{8} \cdot \ln(1/2)\] 4. To find \(t\), divide both sides by \(\ln(1/2)\): \[t = \frac{8\cdot \ln \left(\frac{1}{100}\right)}{\ln(1/2)}\] 5. Finally, calculate the value of \(t\): \[t \approx \frac{8\cdot (-4.605)}{-0.693} \approx 53.26\]
04

Interpret the result

The time it takes for the amount of Iodine-131 to decrease to 1/100th of the original amount is approximately 53.26 days.

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Key Concepts

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

Half-Life
The concept of half-life is critical when discussing radioactive materials like Iodine-131. The half-life of a substance is the time it takes for half of the radioactive atoms in a sample to decay. Over each half-life period, the substance's radioactivity reduces by half.
For Iodine-131, the half-life is 8.0 days. This means if you start with a certain amount, say 10 micrograms, after 8 days, you would have 5 micrograms left. After another 8 days, only 2.5 micrograms would remain.
Understanding this concept is essential in medical and scientific applications, as it influences both dosage and timing for treatment and safety protocols. Essentially, the half-life provides a predictable metric to determine how long a radioactive substance will remain active or potentially hazardous.
  • Predictable decay pattern
  • Used to estimate the longevity of radioactive materials
  • Useful for calculating safe disposal and treatment times
Exponential Decay Formula
The exponential decay formula is a mathematical representation used to describe how the quantity of a radioactive substance decreases over time. The formula is:\( N(t) = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \)where:
- \( N(t) \) is the amount remaining after time \( t \).
- \( N_0 \) is the initial quantity.
- \( T_{1/2} \) is the half-life of the substance.

This formula reveals the exponential nature of decay, meaning the rate of loss decreases over time. That's because each successive half-life reduces the remaining quantity by half, leading to a gradual decrease.
In our exercise, we want to find the time \( t \) when the amount of Iodine-131 decays to \( \frac{1}{100} \) of its original quantity. Plugging the values, we solve the equation, eventually finding \( t \) through steps involving logarithms, highlighting the use of natural logs to simplify exponential equations.

Key points to understand include:
  • Exponential decay leads to a rapid initial decrease
  • Subsequent decreases are less pronounced
  • Logarithms are used to solve for time in equations involving exponential functions
Iodine-131 Applications
Iodine-131 plays a vital role in both the diagnosis and treatment of thyroid-related conditions. Due to its radioactive properties and appropriate half-life, it's particularly useful in medical applications.
In diagnostics, Iodine-131 is used for monitoring thyroid activity through imaging techniques. It helps doctors visualize the gland and assess its functioning by tracking the radiation emitted as the isotope decays.
In treatment, Iodine-131 is used to target and eliminate overactive thyroid tissues or thyroid cancer cells. The isotope delivers localized radiation that interrupts abnormal cell growth while sparing surrounding healthy tissues.

Its effectiveness is due to:
  • Its suitable half-life of 8 days, which balances effectiveness and patient safety
  • The ability to deliver concentrated radiation doses directly to thyroid tissues
  • It offers a non-invasive treatment option for hyperthyroidism and certain thyroid cancers
Understanding these applications showcases the importance of the controlled use of radioactive substances in improving patient outcomes and advancing medical technology.

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Most popular questions from this chapter

The radioactive isotope \(^{242} \mathrm{Cm}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \(^{242} \mathrm{Cm}\) through many transformations to end up as \(^{206} \mathrm{P} \mathrm{b}\) . In the complete decay series, how many \(\alpha\) and \(\beta\) particles are produced?

Using the kinetic molecular theory (section \(5.6 ),\) calculate the root mean square velocity and the average kinetic energy of \(_{1}^{2} \mathrm{H}\) nuclei at a temperature of \(4 \times 10^{7} \mathrm{K}\) . (See Exercise 56 for the appropriate mass values.)

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. 68 Ga (electron capture) b. 62 Cu (positron) c. 212 \(\mathrm{Fr}(\alpha)\) d. 129 \(\mathrm{Sb}(\beta)\)

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction $$ \mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad 8^{\circ}=-2.36 \mathrm{V} $$ a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{G} \circ, \Delta G^{\circ},\) and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred? If \(1.00 \times 10^{3} \mathrm{kg}\) Zreacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at 1.0 \(\mathrm{atm}\) and \(1000 .^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in \(1986,\) hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ It was not possible to prevent a chemical explosion at Chernobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

The stable isotopes of boron are boron-10 and boron-11. Four radioactive isotopes with mass numbers 8, 9, 12, and 13 are also known. Predict possible modes of radioactive decay for the four radioactive isotopes of boron.

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