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Iodine- 125 is often used to treat cancer and has a half-life of 60.1 days. In a sample, the amount of iodine- 125 decreased by \(25 \%\) while in storage. How long was the sample in storage?

Short Answer

Expert verified
The sample was in storage for approximately 24.1 days.

Step by step solution

01

Understand the Problem

We need to determine how long a sample of iodine-125, which has a half-life of 60.1 days, was in storage to experience a 25% decrease in its amount.
02

Identify Key Values

For a decay problem, note that the half-life is 60.1 days. The sample has decreased by 25%, meaning 75% of it remains. This translates to \( \frac{3}{4} \) of the original amount remaining.
03

Use the Exponential Decay Formula

Exponential decay is described by \( N(t) = N_0 \times e^{-\lambda t} \), where \( N(t) \) is the remaining amount, \( N_0 \) is the initial amount, and \( \lambda = \ln(2) / \text{half-life} \).
04

Calculate \( \lambda \)

Calculate the decay constant \( \lambda \) using the half-life: \[ \lambda = \frac{\ln(2)}{60.1} \approx 0.01153 \].
05

Set Up the Equation for the Remaining Amount

Using the formula, set \( \frac{N(t)}{N_0} = \frac{3}{4} \) (since 75% remains): \[ \frac{3}{4} = e^{-\lambda t} \].
06

Solve for \( t \)

Taking the natural logarithm of both sides gives: \[ \ln\left(\frac{3}{4}\right) = -\lambda t \]. Substitute \( \lambda \) and solve for \( t \): \[ t = -\frac{\ln\left(\frac{3}{4}\right)}{0.01153} \approx 24.1 \text{ days} \].
07

Interpret the Solution

It took approximately 24.1 days for the amount of iodine-125 in the sample to decrease by 25% while in storage.

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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 fundamental when dealing with substances undergoing exponential decay, like radioactive isotopes. Half-life is the time it takes for half of a sample to decay. In simpler terms, every half-life, the quantity of the substance is cut in half.
For example, if you start with 100 units of a radioactive material, after one half-life, only 50 units will remain. After two half-lives, you're left with 25 units, and so on. This process continues until the substance has decayed to a negligible amount.
Half-life is crucial in calculating how long it will take for a substance to decrease to a certain level. Understanding this concept allows scientists and medical professionals to determine dosage information, potential exposure times, and how long it takes for a drug or radioactive isotope like Iodine-125 to become less effective or harmful.
Iodine-125
Iodine-125 is a commonly used radioactive isotope in medicine, especially in the treatment of cancer. One of its key features is its relatively short half-life of 60.1 days. This makes it suitable for therapeutic purposes, as it doesn't stay in the body for an extended period.
This isotope is often used in radiotherapy and diagnostic procedures. In cancer treatment, for instance, it's used to target and shrink tumors by emitting radiation directly at the cancer cells, effectively destroying or damaging them so they can’t grow back.
Because of its half-life, there needs to be precise calculation when determining how long Iodine-125 remains active in the body, to ensure safety and effectiveness. Understanding how the isotope decays helps in planning treatment schedules and minimizing exposure risks.
Exponential Decay Formula
The exponential decay formula is crucial for understanding how quantities decrease over time. Particularly in radioactive decay, it models the rate at which an isotope like Iodine-125 loses its radioactivity.
The general formula is:
  • \( N(t) = N_0 \times e^{-\lambda t} \)
Here, \( N(t) \) represents the amount of the substance remaining after time \( t \. \), \( N_0 \) is the initial amount, and \( \lambda \) is the decay constant. This constant \( \lambda \) is derived from the half-life, calculated as \( \lambda = \frac{\ln(2)}{\text{half-life}} \).
By using this formula, you can determine how long it takes for a certain percentage of the substance to decay. In the problem at hand, it allowed us to find the time it took for a 25% decrease in Iodine-125, helping to round out the effective use and disposal management of radioactive materials.

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