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Chapter 18: Problem 34

Radioisotopes of iodine are widely used in medicine. For example, iodine-131 \(\left(t_{1 / 2}=8.04\right.\) days) is used to treat thyroid cancer. If you ingest a sample of \(\mathrm{Na}^{131} \mathrm{I},\) calculate the time required for the isotope to decrease to \(5.0 \%\) of its original activity.

Short Answer

Expert verified
The time required is approximately 35.8 days.

Step by step solution

01

Understanding Half-Life

The half-life of iodine-131 is 8.04 days. This means that every 8.04 days, half of the iodine-131 decays.
02

Recognizing the Exponential Decay Equation

The activity of a radioactive substance is given by the equation \( A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \), 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

Setting Up the Equation for 5% Activity

To find when the activity is 5% of the original, set \( A = 0.05 A_0 \). Substitute into the equation: \( 0.05 A_0 = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{8.04}} \). Cancel \( A_0 \) from both sides to get \( 0.05 = \left( \frac{1}{2} \right)^{\frac{t}{8.04}} \).
04

Solving the Exponential Equation

Take the logarithm of both sides: \( \ln(0.05) = \ln\left( \left( \frac{1}{2} \right)^{\frac{t}{8.04}} \right) \). This simplifies to \( \ln(0.05) = \frac{t}{8.04} \times \ln\left( \frac{1}{2} \right) \).
05

Isolating \(t\)

Solve for \( t \) by rearranging the equation: \( t = \frac{8.04 \times \ln(0.05)}{\ln(\frac{1}{2})} \).
06

Calculating \(t\)

Use a calculator to compute \( t = \frac{8.04 \times \ln(0.05)}{\ln(0.5)} \). The calculated time \( t \) is approximately 35.8 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 key to understanding radioactive decay. It represents the time required for half of a radioactive substance to decay. For iodine-131, the half-life is 8.04 days. This means that every 8.04 days, the amount of iodine-131 decreases by half.
Half-life is a constant value specific to each radioactive material. It doesn't depend on how much substance you start with.
  • If you start with 10 units, after one half-life, you'll have 5 units left.
  • After two half-lives, you'll have 2.5 units left, and so on.
Understanding half-life helps predict how long it will take for a substance to become either safe or ineffective.
Iodine-131
Iodine-131 is a radioactive isotope of iodine. It's commonly used in medical applications, particularly in treating thyroid cancer. Due to its radioactive nature, iodine-131 emits particles that can destroy diseased thyroid tissues.
This isotope is carefully administered in controlled doses to target and treat specific areas in the thyroid gland.
  • Its half-life of 8.04 days is significant enough to allow for effective treatment over a set period.
  • Post-treatment, the remaining iodine-131 breaks down, reducing radiation exposure over time.
Working with iodine-131 requires understanding its half-life to ensure it is used effectively and safely.
Exponential Decay
Exponential decay is a process where the quantity decreases at a rate proportional to its current value. Radioactive substances like iodine-131 follow this pattern. The decay formula used to track remaining radioactive activity is:
\[ A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where:
  • \( A \) is the remaining activity at time \( t \).
  • \( A_0 \) is the initial activity.
  • \( t \) is the elapsed time.
  • \( t_{1/2} \) is the half-life.
This formula shows how each period (such as 8.04 days for iodine-131) reduces the activity by half. For the case where activity needs to decline to 5%, you set \( A = 0.05 A_0 \) and solve for \( t \). This calculation shows how long it takes for the substance to reduce to a specific amount of its original value. Understanding exponential decay is essential to predict the long-term behavior of radioactive materials.

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Most popular questions from this chapter

All radioactive decays are first order. Why is this so?

Radioactive isotopes are often used as "tracers" to follow an atom through a chemical reaction. Acetic acid reacts with methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), by eliminating a molecule of \(\mathrm{H}_{2} \mathrm{O}\) to form methyl acetate, \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\). Explain how you would use the radioactive isotope \({ }^{18} \mathrm{O}\) to show whether the oxygen atom in the water product comes from the \(-\mathrm{OH}\) of the acid or the \(-\mathrm{OH}\) of the alcohol. $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{OH} \longrightarrow \mathrm{CH}_{3} \mathrm{COOCH}_{3}+\mathrm{H}_{2} \mathrm{O}\\\ &\begin{array}{lll} \text { acetic acid } & \text { methanol } & \text { methyl acetate } \end{array} \end{aligned} $$

In June 2009 , researchers at the Joint Institute of Nuclear Research in Dubna, Russia, began continuous bombardment of a \(22-\mathrm{mg}\) target of berkelium- 249 atoms with calcium- 48 nuclei in an attempt to synthesize element 117 . The bombardment continued until January 2010 when evidence indicated that a few atoms of the new element had been synthesized. The half- life of berkelium- 249 is 330 days. Assuming that the reaction period was 240 days, calculate the mass of berkelium-249 remaining.

During the Three Mile Island incident, people in central Pennsylvania were concerned that strontium-90 (a beta emitter) released from the reactor could become a health threat (it did not). Where would this isotope collect in the body? If so, what types of problems could it cause?

A sample of the alpha emitter \({ }_{86}^{222} \mathrm{Rn}\) had an initial activity, \(A_{0},\) of \(7.00 \times 10^{4}\) Bq. After 10.0 days its activity, \(A\) had fallen to \(1.15 \times 10^{4} \mathrm{~Bq} .\) Calculate the decay constant and half-life of radon-222.
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