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Chapter 31: Problem 35

Strontium \({ }_{38}^{90} \mathrm{Sr}\) has a half-life of 29.1 yr. It is chemically similar to calcium, enters the body through the food chain, and collects in the bones. Consequently, \({ }_{38}^{90} \mathrm{Sr}\) is a particularly serious health hazard. How long (in years) will it take for \(99.9900 \%\) of the \({ }_{38}^{90} \mathrm{Sr}\) released in a nuclear reactor accident to disappear?

Short Answer

Expert verified
It will take approximately 387 years for 99.9900% of Strontium-90 to decay.

Step by step solution

01

Understanding the Problem

We need to find out how long it takes for 99.9900% of Strontium-90 (\(^{90}{\rm{Sr}}\)) to decay. This means only 0.01% of \(^{90}{\rm{Sr}}\) should remain. We will use the concept of half-life and exponential decay to solve this.
02

Setting Up the Formula

The formula to calculate the remaining quantity after time \(t\) when the half-life is known is: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]where \(N(t)\) is the remaining quantity, \(N_0\) is the initial quantity, \(t\) is the time elapsed, and \(T_{1/2}\) is the half-life of the isotope.
03

Calculating the Remaining Fraction

We want to find when 0.01% of the initial quantity remains. Therefore, \[ \frac{N(t)}{N_0} = 0.0001 \]
04

Equation Relating Time and Half-Life

Substituting into the decay formula, we get:\[ 0.0001 = \left(\frac{1}{2}\right)^{\frac{t}{29.1}} \]To solve for \(t\), we will use logarithms.
05

Applying Logarithms

Taking the logarithm of both sides, we have:\[ \log(0.0001) = \log \left(\left(\frac{1}{2}\right)^{\frac{t}{29.1}}\right) \]Using the logarithm power rule, this becomes:\[ \log(0.0001) = \frac{t}{29.1} \cdot \log\left(\frac{1}{2}\right) \]
06

Solving for Time \(t\)

Isolating \(t\), we have:\[ t = \frac{\log(0.0001)}{\log\left(\frac{1}{2}\right)} \times 29.1 \]Calculating this gives:\[ t \approx \frac{-4}{-0.3010} \times 29.1 \approx 386.85 \]
07

Rounding the Result

Therefore, it will take approximately 386.85 years for 99.9900% of the Strontium-90 to decay.

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Key Concepts

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

Strontium-90
Strontium-90 is a radioactive isotope that is similar to calcium in its chemical behavior. Because of this similarity, Strontium-90 can enter the body through the food chain and accumulate in the bones, posing significant health risks. It is one of the by-products of nuclear fission and can be released during nuclear reactor accidents. This makes understanding its decay process critical for assessing long-term exposure risks.
Half-life
The half-life of a radioactive substance is the time it takes for half of the material to decay. For Strontium-90, this period is 29.1 years. This value remains constant, no matter how much substance you start with.
  • After the first 29.1 years, 50% of Strontium-90 remains.
  • After another 29.1 years (total 58.2 years), 25% remains.
The concept of half-life helps predict how quickly a radioactive isotope will decay over time, which is essential for calculating safety and exposure levels.
Exponential Decay
Exponential decay occurs when the rate of a quantity's decrease is proportional to its current value. Radioactive substances, like Strontium-90, decay in this manner.
The formula for exponential decay is:\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]where:
  • \( N(t) \) is the amount remaining after time \( t \).
  • \( N_0 \) is the initial amount.
  • \( T_{1/2} \) is the half-life.
This formula helps us understand how quickly a substance, like Strontium-90, decreases over time, allowing us to evaluate the time required for significant decay.
Logarithmic calculations
Logarithmic calculations allow us to solve for time \( t \) in exponential decay problems. By taking logarithms of both sides of the decay formula, it's possible to isolate \( t \). For instance, in finding how long it takes for 99.9900% of Strontium-90 to decay, we start with:
\[ 0.0001 = \left(\frac{1}{2}\right)^{\frac{t}{29.1}} \]By applying logarithms:
  • \[ \log(0.0001) = \frac{t}{29.1} \cdot \log\left(\frac{1}{2}\right) \]
  • Solving for \( t \), we get:
    \[ t = \frac{\log(0.0001)}{\log\left(\frac{1}{2}\right)} \times 29.1 \]
Calculating this gives approximately 386.85 years. These calculations allow for precise predictions of decay durations.

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Most popular questions from this chapter

A sample has a \(_{6}^{14} \mathrm{C}\) activity of \(0.0061 \mathrm{Bq}\) per gram of carbon. (a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq. (b) Evidence suggests that the value of \(0.23 \mathrm{Bq}\) might have been as much as \(40 \%\) larger. Repeat part (a), taking into account this \(40 \%\) increase.

Suppose that you could pack neutrons (mass \(\left.=1.67 \times 10^{-27} \mathrm{kg}\right)\) inside a tennis ball (radius \(=0.032 \mathrm{m}\) ) in the same way as neutrons and protons are packed together in the nucleus of an atom. (a) Approximately how many neutrons would fit inside the tennis ball? (b) A small object is placed \(2.0 \mathrm{m}\) from the center of the neutron-packed tennis ball, and the tennis ball exerts a gravitational force on it. When the object is released, what is the magnitude of the acceleration that it experiences? Ignore the gravitational force exerted on the object by the earth.

Complete the following decay processes by stating what the symbol \(X\) represents \(\left(X=\alpha, \beta^{-}, \beta^{+},\right.\) or \(\left.\gamma\right)\): (a) \({ }_{82}^{211} \mathrm{Pb} \rightarrow{ }_{83}^{211} \mathrm{Bi}+\mathrm{X}\) (b) \({ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{B}+\mathrm{X}\) (c) \(_{90 }^{231} \mathrm{Th}^{*} \rightarrow{ }_{90}^{231} \mathrm{Th}+\mathrm{X}\) (d) \(^{210}_{84}\mathrm{Po} \rightarrow{ }_{82}^{206} \mathrm{Pb}+\mathrm{X}\)

By what factor does the nucleon number of a nucleus have to increase in order for the nuclear radius to double?

The \({ }_{15}^{32} \mathrm{P}\) isotope of phosphorus has a half-life of 14.28 days. What is its decay constant in units of \(s^{-1} ?\)
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