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Iodine-131 is used as a nuclear medicine to treat hyperthyroidism. The half- life of \({ }^{131}\) I is 8.04 days. How long will it take for a \(500 \mathrm{mg}\) sample of \({ }^{131}\) I to decay into \(1 \%\) of its original mass?

Short Answer

Expert verified
It takes approximately 53.47 days for the sample to decay to 1% of its original mass.

Step by step solution

01

Understand Half-life Concept

The half-life of a substance is the time it takes for half of the sample to decay. In this case, the half-life of Iodine-131 is 8.04 days.
02

Set Up the Decay Formula

Use the exponential decay formula \( N = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \), where \( N \) is the remaining amount, \( N_0 \) is the initial amount, \( t \) is the time, and \( T_{1/2} \) is the half-life.
03

Define Variables

Let \( N_0 = 500 \) mg (initial mass) and \( N = 0.01 \times 500 = 5 \) mg (1% of initial mass). The half-life, \( T_{1/2} \), is 8.04 days.
04

Insert Values into the Formula

Substitute the values into the formula to get \( 5 = 500 \cdot \left(\frac{1}{2}\right)^{\frac{t}{8.04}} \).
05

Solve for \( t \)

First, divide both sides by 500: \( 0.01 = \left(\frac{1}{2}\right)^{\frac{t}{8.04}} \). Taking the natural logarithm of both sides, \( \ln(0.01) = \frac{t}{8.04} \cdot \ln(0.5) \).
06

Calculate \( t \)

Rearrange to solve for \( t \): \( t = \frac{\ln(0.01)}{\ln(0.5)} \times 8.04 \). Compute this to find \( t \).
07

Perform Calculation

Calculate \( t \approx \frac{(-4.60517)}{-0.693147} \times 8.04 \approx 53.47 \). Thus, it takes approximately 53.47 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 a crucial part of understanding radioactive decay. Half-life refers to the time it takes for half of the radioactive atoms in a sample to decay. It helps predict how quickly a substance will decrease and is vital in fields like nuclear medicine and geology.

For Iodine-131, a commonly used radioactive isotope in medical treatments, the half-life is 8.04 days. This means every 8.04 days, half of the Iodine-131 present will have decayed into another element. Over two half-lives, only a quarter of the original amount would remain, and so on.
  • This steady and predictable rate allows scientists and doctors to plan treatments and ensure patient safety.
  • Knowing the half-life helps healthcare professionals determine the duration of radiation exposure a patient might experience.
Understanding half-life is not just about measuring time; it’s about understanding the natural decay process and how long a material remains active and effective.
Exponential Decay Formula
The exponential decay formula is a mathematical representation used to describe how the amount of a radioactive substance decreases over time. The formula is expressed as:\[ N = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]Where:
  • \( N \) is the remaining quantity of the substance.
  • \( N_0 \) is the initial quantity.
  • \( t \) represents the total time elapsed.
  • \( T_{1/2} \) is the half-life of the substance.
This formula shows how over each half-life, the substance's mass decreases by half.

To solve problems involving radioactive decay, such as determining how long it will take for a sample to decay to a certain percentage of its original mass, this formula is invaluable. By manipulating the formula, you can find out how long the decay will take or determine how much of a substance remains after a given time.
Using logarithms helps us solve for \( t \) when juggling with powers of fractions, making these calculations both precise and practical.
Iodine-131
Iodine-131 is a radioactive isotope of iodine that plays an essential role in the field of nuclear medicine. Discovered in the 1930s, Iodine-131 emits both beta and gamma radiation, which makes it particularly useful for medical applications.

In medicine, Iodine-131 is commonly used to treat conditions such as hyperthyroidism and certain types of thyroid cancer. It's preferred in these treatments because of its half-life of about 8.04 days, allowing it to deliver therapeutic doses effectively while minimizing patient exposure to radiation over the long term.
  • The radiation from Iodine-131 targets thyroid cells effectively, either shrinking or eliminating problematic tissues.
  • Its properties make it both a therapy and a diagnostic tool, as it can be traced within the body using various imaging technologies.
This dual capacity ensures it remains a cornerstone in treating thyroid-related conditions, providing both critical information and effective means to combat disease.
Nuclear Medicine
Nuclear medicine is a medical specialty that uses radioactive substances for diagnosis and therapy. Unlike traditional medical imaging that shows structure, nuclear medicine allows for the visualization of physiological functions. This technique provides not just images but insightful data about how organs and tissues function.

Radioactive isotopes, like Iodine-131, are used in various treatments and diagnostics. These substances are chosen for their ability to emit either beta or gamma rays, which can be detected by special imaging devices.
  • Nuclear medicine techniques include treatments for cancer, heart disease, and other serious conditions.
  • It provides critical information that often can't be obtained through other imaging techniques, offering a deeper look into bodily functions.
The reach of nuclear medicine goes beyond treatment, offering physicians a dynamic tool for management and diagnosis, leading to personalized and highly effective care for patients.

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Most popular questions from this chapter

The nuclear masses of \({ }^{7} \mathrm{Be},{ }^{9} \mathrm{Be},\) and \({ }^{10} \mathrm{Be}\) are 7.0147,9.0100 , and \(10.0113 \mathrm{u}\), respectively. Which of these nuclei has the largest binding energy per nucleon?

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .(\mathbf{a})\) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left(\begin{array}{c}235 \\ \mathrm{U} \text { nuclear mass, } 234.9935 \mathrm{u} ;{ }^{141} \mathrm{Ba} \text { nuclear mass, }\end{array}\right.\) \(140.8833 \mathrm{u} ;{ }^{92} \mathrm{Kr}\) nuclear mass, \(91.9021 \mathrm{u}\) ) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in 1.0 day?

A radioactive decay series that begins with \({ }_{90}^{232}\) Th ends with formation of the stable nuclide \({ }_{82}^{208} \mathrm{~Pb} .\) How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Charcoal samples from Stonehenge in England were burned in \(\mathrm{O}_{2}\), and the resultant \(\mathrm{CO}_{2}\) gas bubbled into a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (limewater), resulting in the precipitation of \(\mathrm{CaCO}_{3}\). The \(\mathrm{CaCO}_{3}\) was removed by filtration and dried. A 788-mg sample of the \(\mathrm{CaCO}_{3}\) had a radioactivity of \(1.5 \times 10^{-2} \mathrm{~Bq}\) due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half- life of carbon-14, 5700 yr, calculate the age of the charcoal sample.

Write balanced equations for each of the following nuclear reactions: \((\mathbf{a}){ }_{92}^{238} \mathrm{U}(\mathrm{n}, \gamma){ }^{239} \mathrm{U},(\mathbf{b}){ }_{8}^{16} \mathrm{O}(\mathrm{p}, \alpha){ }^{13} \mathrm{~N},\) (c) \({ }_{8}^{18} \mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }_{9}^{19} \mathrm{~F} .\)

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