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Chapter 7: Problem 44

The radioactive isotope Polonium-214 has a half-life of \(0.00014\) seconds. Suppose 1 kilogram of the polonium appears in your sink. a. Find the values of \(a\) and \(b\) such that the amount in grams after \(t\) seconds is given by \(f(t)=a\left(\frac{1}{2}\right)^{b t}\). b. After a \(\frac{1}{100}\) second blink, would a gram of polonium remain? Explain your answer.

Short Answer

Expert verified
No, less than a gram remains; \( t = \frac{1}{100} \) causes significant decay.

Step by step solution

01

Understand the Exponential Decay Formula

The exponential decay of a substance is generally expressed as \( f(t) = a \left( \frac{1}{2} \right)^{bt} \). Here, \( a \) is the initial amount of the substance, and \( b \) is a decay constant. We need to find these values based on the given half-life.
02

Establish the Initial Amount for Polonium-214

The problem states that we start with 1 kilogram of polonium, but since the equation needs grams, convert kilograms to grams. Hence, \( a = 1000 \) grams.
03

Determine the Decay Constant (b) Using Half-life

The half-life of Polonium-214 is given as \(0.00014\) seconds. For half-life \( t_{1/2} \), \( f(t_{1/2}) = \frac{a}{2} \). So, the equation becomes: \[ 1000 \left( \frac{1}{2} \right)^{b \times 0.00014} = 500 \]This simplifies to:\[ \left( \frac{1}{2} \right)^{0.00014b} = \frac{1}{2} \]Taking the logarithm of both sides gives:\[ 0.00014b \cdot \log\left(\frac{1}{2}\right) = \log\left(\frac{1}{2}\right) \]Solving for \( b \), we get \( b = \frac{1}{0.00014} \).
04

Solve for b

From the equation in Step 3, solve:\[ b = \frac{1}{0.00014} \approx 7142.86 \]Thus, \( b \approx 7142.86 \).
05

Substitute a and b into the Decay Formula

Substitute \( a = 1000 \) and \( b = 7142.86 \) into the formula:\[ f(t) = 1000 \left( \frac{1}{2} \right)^{7142.86t} \]
06

Check Polonium Amount After a Blink

Now check if more than a gram remains after \( t = \frac{1}{100} \) seconds:\[ f\left( \frac{1}{100} \right) = 1000 \left( \frac{1}{2} \right)^{7142.86 \times \frac{1}{100}} \]Simplify the exponent:\[ = 1000 \left( \frac{1}{2} \right)^{71.4286} \]Since \( \left( \frac{1}{2} \right)^{71.4286} \) is a very small number, the final result is much less than 1 gram.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a process where unstable atomic nuclei lose energy by emitting radiation. Over time, this results in the transformation of the nucleus into a different element or a different isotope of the same element. This decay occurs naturally and is characterized by a decrease in the quantity of the radioactive substance, which we describe using the concept of exponential decay.
This type of decay is unpredictable for a single unstable nucleus, but statistically, it follows a pattern that can be predicted for a larger number of nuclei. Radioactive decay is critical for understanding processes like dating archaeological artifacts and managing nuclear materials safely.
Half-Life
The half-life of a radioactive substance is the time it takes for half of the original amount of the substance to decay. It's a fundamental parameter in radioactive decay and provides insight into the rate of decay.
For instance, if we start with 1 kilogram of a substance with a half-life of 0.00014 seconds, after that duration, only half of the substance will remain.
Importantly, half-life does not change. Even after several half-lives, each period sees half of the remaining substance decayed. This constancy makes it a valuable measure for scientists. In the exercise, understanding the half-life of Polonium-214 helped determine how quickly the material decayed.
Decay Constant
The decay constant (\(b\)) is a crucial part of understanding the exponential decay equation: \(f(t) = a \left(\frac{1}{2}\right)^{bt}\). It's a constant that helps determine how fast a substance decays.
  • The constant itself is derived from the half-life, using the relationship \(b = \frac{1}{t_{1/2}}\).
  • This relationship shows that the decay constant is inversely proportional to the half-life; a shorter half-life means a larger decay constant, indicating faster decay.
  • In the case of Polonium-214 from the exercise, the very short half-life yielded a high decay constant of approximately 7142.86, meaning it decays very rapidly.
Exponential Functions
Exponential functions are mathematical expressions that model situations where a quantity grows or decays at a rate proportional to its current value. The general form is \(f(t) = a b^t\).
In radioactive decay, specifically, the base of the exponential function is \(\frac{1}{2}\), reflecting how the substance halves over each half-life period.

Characteristics of exponential functions include:
  • An initial amount (\(a\)) that serves as the starting value from which decay or growth occurs.
  • A base (\(b\)), which in decay processes such as radioactive decay, is \(\frac{1}{2}\), showing how quantities reduce.
  • A variable exponent \(t\) that represents time, showing how the quantity changes over time.
Understanding these functions is key to analyzing and predicting the behavior of various processes, including the decay detailed in the exercise.

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