/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 The exercises in this set are gr... [FREE SOLUTION] | 91Ó°ÊÓ

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The exercises in this set are grouped according to discipline. They involve exponential or logarithmic models. An initial amount of a radioactive substance \(y_{0}\) is given, along with information about the amount remaining after a given time t in appropriate units. For an equation of the form \(y=y_{0} e^{k t}\) that models the situation, give the exact value of \(k\) in terms of natural logarithms. \(y_{0}=10 \mathrm{mg} ;\) The half-life is 100 days.

Short Answer

Expert verified
The decay constant \( k \) is \( \frac{\ln(0.5)}{100} \)

Step by step solution

01

- Write the exponential decay formula

The exponential decay formula is given by the equation: \[ y = y_{0} e^{k t} \] where \( y \) is the amount of substance remaining, \( y_{0} \) is the initial amount of substance, \( k \) is the decay constant, and \( t \) is the time.
02

- Apply the known values

We know that the initial amount \( y_{0} = 10 \text{ mg} \), and after 100 days (which is half-life), half of the initial amount remains. Thus, \( y = \frac{y_{0}}{2} = 5 \text{ mg} \) and \( t = 100 \) days.
03

- Substitute the values into the formula

Substitute \( y = 5 \text{ mg} \), \( y_{0} = 10 \text{ mg} \), and \( t = 100 \) into the equation: \[ 5 = 10 e^{k \times 100} \]
04

- Solve for the decay constant \( k \)

First divide both sides of the equation by 10 to isolate the exponential term: \[ \frac{5}{10} = e^{k \times 100} \]\[ 0.5 = e^{100k} \]Take the natural logarithm of both sides to solve for \( k \): \[ \ln(0.5) = \ln(e^{100k}) \]By the property of logarithms, this simplifies to: \[ \ln(0.5) = 100k \]Finally, solve for \( k \): \[ k = \frac{\ln(0.5)}{100} \]

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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 the process by which an unstable atomic nucleus loses energy by emitting radiation. This decay occurs at an exponential rate, meaning it decreases by a constant proportion in each time period. This process can be described using an exponential decay formula:
  • \[y = y_{0} e^{k t}\]
  • Where:
  • \(y\) is the amount of substance remaining
  • \(y_{0}\) is the initial amount of substance
  • \(k\) is the decay constant
  • \(t\) is the time
Understanding how radioactive decay works allows us to predict the amount of substance that will be left after a given period, which is useful in fields like nuclear physics, medicine, and archaeology.
Natural Logarithms
Natural logarithms (ln) are a type of logarithm where the base is the mathematical constant e (approximately 2.71828). They are particularly useful in the context of exponential growth and decay equations. The properties of natural logarithms make it easier to solve equations where the variable is an exponent, such as in the exponential decay formula.
For example, when solving for the decay constant in the formula \(y = y_{0} e^{k t}\), taking the natural logarithm of both sides allows us to isolate the variable \(k\):
  • \[0.5 = e^{100k}\]
  • Taking the natural logarithm of both sides gives:
  • \[ln(0.5) = ln(e^{100k})\]
  • Using the property \(ln(a^b) = b\cdot ln(a)\):
  • \[ln(0.5) = 100k\]
This step is crucial for separating the variable we need to solve for.
Half-Life Calculation
The half-life of a radioactive substance is the time it takes for half of the substance to decay. It's a fundamental concept because it provides a measure of how quickly a substance undergoes radioactive decay. The relationship between the half-life and the decay constant is given by the exponential decay formula.
For instance, if we know the half-life (in this case, 100 days), we can determine how much of the substance remains after a certain period. With our initial amount \(y_{0} = 10 \text{ mg}\)and half-life of 100 days, we know that in 100 days, the amount remaining will be \(\frac{10}{2} = 5 \text{ mg}\).
Using the formula again, we have
  • \[5 = 10 e^{k \times 100}\]
  • Dividing both sides by 10:
  • \[0.5 = e^{100k}\]
This helps in solving the equation for the decay constant.
Decay Constant
The decay constant (\(k\))is a crucial value in the exponential decay formula. It represents the rate at which a substance decays. A higher decay constant means the substance decays faster. The decay constant can be found using natural logarithms once we've established the half-life or remaining amount after a certain time. To find the decay constant from our example:
  • Substitute the known values into the formula:
  • \[0.5 = e^{100k}\]
  • Take the natural logarithm of both sides:
  • \[ln(0.5) = ln(e^{100k})\]
  • Simplify using the property \(ln(a^b) = b\cdot ln(a)\):
  • \[ln(0.5) = 100k\]
  • Solving for the decay constant:
  • \[k = \frac{ln(0.5)}{100}\]
This constant helps in predicting how much of a substance will remain after any given period.

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Most popular questions from this chapter

Emissions Tax One action that government could take to reduce carbon emissions into the atmosphere is to levy a tax on fossil fuel. This tax would be based on the amount of carbon dioxide emitted into the air when the fuel is burned. The cost-benefit equation $$\ln (1-P)=-0.0034-0.0053 T$$ models the approximate relationship between a tax of \(T\) dollars per ton of carbon and the corresponding percent reduction \(P\) (in decimal form) of emissions of carbon dioxide. (Source: Nordhause, W., "To Slow or Not to Slow: The Economics of the Greenhouse Effect," Yale University, New Haven, Connecticut.) (a) Write \(P\) as a function of \(T\). (b) Graph \(P\) for \(0 \leq T \leq 1000 .\) Discuss the benefit of continuing to raise taxes on carbon (c) Determine \(P\) when \(T=60\) dollars, and interpret this result. (d) What value of \(T\) will give a \(50 \%\) reduction in carbon emissions?

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