The half life of \(_{38} \mathrm{Sr}^{90}\) is 20 year. If the sample of this nucleide has an activity of 8,000 disintegrations \(\min ^{-1}\) today, what will be its activity after 80 year?

Short Answer

Expert verified
The activity after 80 years will be 500 disintegrations per minute.

Step by step solution

01

Understand Half-Life

Half-life is the amount of time required for a quantity to reduce to half its initial value. For a substance with a half-life of 20 years, the quantity will halve every 20 years.
02

Calculate the Number of Half-Lives

To find the number of half-lives that pass in 80 years for a substance with a 20-year half-life, divide the total time elapsed by the half-life duration: Number of Half-Lives = Total Time Elapsed / Half-Life Duration.
03

Determine the Remaining Activity

Use the formula for exponential decay, which is Final Activity = Initial Activity * (1/2)^(Number of Half-Lives), substituting the values for the initial activity and the number of half-lives calculated in the previous step.

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Key Concepts

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

Half-Life Calculation
Understanding half-life is essential when studying radioactive materials. The half-life of a radioactive substance is the time required for half of the atoms in a sample to decay or transform to a different nuclide.

This concept is crucial for multiple fields, from archaeology to nuclear physics. It gives scientists a way to date ancient objects, visualize the safety of nuclear waste storage timelines, and calculate medication dosages in medical treatments.

To calculate the half-life, one typically uses the formula: \[\text{Number of Half-Lives} = \frac{\text{Total Time Elapsed}}{\text{Half-Life Duration}}\].For example, with a half-life of 20 years, after 80 years, a substance would have experienced \[\text{Number of Half-Lives} = \frac{80 \text{ years}}{20 \text{ years}} = 4\] half-lives. This illustrates that the original sample would have halved 4 times, subsequently reducing its initial quantity to one-sixteenth of its original amount.
Exponential Decay Formula
Radioactive decay follows an exponential decay model, which means that it decreases at a rate proportional to its current value. This leads to a rapid drop in quantity initially, which slows down as time goes on but never truly hits zero.

The formula for exponential decay is given by \[\text{Final Activity} = \text{Initial Activity} \times \left(\frac{1}{2}\right)^{\text{Number of Half-Lives}}\].

Applying the Exponential Decay Formula

Using the provided problem as an example, with an initial activity of 8,000 disintegrations per minute and a half-life of 20 years, after 80 years (which is 4 half-lives):\[\text{Final Activity} = 8000 \times \left(\frac{1}{2}\right)^4\].By performing the calculation, we find that the activity after 80 years would be 500 disintegrations per minute. The use of the exponential decay formula is pivotal in accurately determining future activity levels in radioactive substances.
Activity of a Radioactive Sample
The activity of a radioactive sample is defined as the rate at which the atoms within the sample decay. It is generally expressed in terms of disintegrations per unit time, for example, disintegrations per minute or per second.

Activity can provide insights into the stability of a radioactive sample, its potential for harmful radiation, and its usefulness in a variety of applications, such as medical imaging. The initial activity is a snapshot of this rate at a starting point, which will change over time as the sample decays.

Factors Affecting Activity

Several factors can affect the activity of a given sample, including:
  • The type of radioactive isotope
  • The original number of radioactive atoms present
  • The half-life of the isotope
By understanding these factors, one can better predict how the activity of a sample will change over time and how it can impact its surrounding environment or its use in various applications. For instance, a short half-life indicates a high initial activity that falls off rapidly, whereas a long half-life implies a more stable sample with activity declining slowly over a prolonged period.

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Most popular questions from this chapter

Give one example each of (a) \(\alpha\) -emission, (b) \(\beta^{+}\) -emission, and (c) K-capture. Write the equation for these nuclear changes.

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