/*! 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 19 One of the main contaminants of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a rate of approximately \(2.5 \%\) per year. (a) Write the percent of strontium-90 remaining, \(P\), as a function of years, \(t\), since the nuclear accident. [Hint: \(100 \%\) of the contaminant remains at \(t=0 .]\) (b) Graph \(P\) against \(t\) (c) Estimate the half-life of strontium-90. (d) After the Chernobyl disaster, it was predicted that the region would not be safe for human habitation for 100 years. Estimate the percent of original strontium-90 remaining at this time.

Short Answer

Expert verified
(a) \( P(t) = 100(0.975)^t \); (c) Half-life \(\approx 27.72\) years; (d) \(\approx 7.72\%\) remains after 100 years.

Step by step solution

01

Understanding the Exponential Decay Formula

The decay process can be modeled using the exponential decay formula \( P(t) = P_0 e^{-kt} \), where \( P(t) \) is the percentage of the substance remaining at time \( t \), \( P_0 \) is the initial amount, \( k \) is the decay constant, and \( e \) is the base of the natural logarithm.
02

Determining the Decay Constant

The decay rate is given as 2.5% per year. To find \( k \), the decay constant, we use the formula \( k = \ln(1 - \, ext{decay rate}) \). Converting 2.5% to a decimal gives 0.025, so \( k = \ln(1 - 0.025) = \ln(0.975) \).
03

Writing the Function for Strontium-90 Decay

Substitute \( P_0 = 100 \) (since 100% is the initial amount at \( t=0 \)) and the calculated \( k \) into the decay formula: \[ P(t) = 100e^{- ext{ln}(0.975)t} = 100(0.975)^t \].
04

Graphing the Function

To graph \( P \) against \( t \), plot the exponential decay function \( P(t) = 100(0.975)^t \) for \( t \) ranging from 0 to 100 years. This will show a decreasing curve starting at 100 at \( t=0 \) and approaching zero as \( t \) increases.
05

Estimating the Half-Life of Strontium-90

The half-life is the time it takes for half of the substance to decay. Set \( P(t) = 50 \) and solve \( 50 = 100(0.975)^t \). Simplify to \( 0.5 = (0.975)^t \). Taking the natural logarithm of both sides, \( t \approx \frac{\ln(0.5)}{\ln(0.975)} \approx 27.72 \) years.
06

Estimating the Remaining Percentage After 100 Years

Substitute \( t = 100 \) into the decay formula: \( P(100) = 100(0.975)^{100} \). Calculate this to find \( P(100) \approx 7.72 \% \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Half-life calculation
Half-life is a fundamental concept in understanding exponential decay, especially in the context of radioactive materials. Simply put, half-life is the time required for a quantity to reduce to half its initial amount. In real-world terms, it's how long it takes for half of a radioactive substance to undergo decay and turn into something else.

For strontium-90, which is a byproduct of nuclear reactions, the half-life can be calculated using the formula for exponential decay. The formula is:
  • \[ P(t) = P_0 (0.975)^t \]
This represents the percentage of strontium-90 remaining after time \( t \), given a decay rate of 2.5\% per year.

To find the half-life, you set \( P(t) = 50 \) in this formula, because you are looking for the time when only 50\% of the original amount is left. By solving the equation \[ 0.5 = (0.975)^t \] using logarithms, you find that the half-life of strontium-90 is approximately 27.72 years. This means after this time frame, half of the initial strontium-90 has decayed.
Radioactive decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. In the case of strontium-90, as it decays, it transforms into another element. This process follows an exponential decay model.This model is represented by the equation \[ P(t) = P_0 e^{-kt} \], where \( P(t) \) is the remaining substance at time \( t \), \( P_0 \) is the initial amount, and \( k \) is the decay constant.

The decay constant is found using \[ k = ext{ln}(1 - ext{decay rate}) \]. For strontium-90, with a decay rate of 2.5\%, you calculate \( k = ext{ln}(0.975) \).

This constant \( k \) shows how quickly or slowly the substance is decaying. A higher value of \( k \) would indicate a faster decay process. Understanding this constant is crucial for making accurate predictions about the material's future activity.
Graphing exponential functions
Graphing exponential functions provides a visual representation of how quantities change over time. In the case of radioactive decay, such as the strontium-90 example, graphing provides insights into how the remaining percentage of a substance decreases.The function \[ P(t) = 100(0.975)^t \] describes the decay over time, starting completely intact at 100\% (\( P_0 = 100 \)). When graphed, this function forms a downward-sloping curve which gets closer to zero but never actually reaches it as time progresses.

Plotting \( P \) against \( t \) from time \( t=0 \) to 100 years, you would observe:
  • A rapid decline at the start of the curve
  • A slowing rate of decline as \( t \) increases
  • After about 100 years, as per calculations, a small portion (around 7.72\%) of the original substance remains
This graphing approach helps illustrate the concept of half-life and the nature of exponential decay, making it easier to comprehend how radioactive substances diminish over time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A person breathes in and out every three seconds. The volume of air in the person's lungs varies between a minimum of 2 liters and a maximum of 4 liters. Which of the following is the best formula for the volume of air in the person's lungs as a function of time? (a) \(y=2+2 \sin \left(\frac{\pi}{3} t\right)\) (b) \(y=3+\sin \left(\frac{2 \pi}{3} t\right)\) (c) \(y=2+2 \sin \left(\frac{2 \pi}{3} t\right)\) (d) \(y=3+\sin \left(\frac{\pi}{3} t\right)\)

A health club has cost and revenue functions given by \(C=10,000+35 q\) and \(R=p q\), where \(q\) is the number of annual club members and \(p\) is the price of a oneyear membership. The demand function for the club is \(q=3000-20 p\) (a) Use the demand function to write cost and revenue as functions of \(p\). (b) Graph cost and revenue as a function of \(p\), on the same axes. (Note that price does not go above $$\$ 170$$ and that the annual costs of running the club reach \(\$ 120,000 .)\) (c) Explain why the graph of the revenue function has the shape it does. (d) For what prices does the club make a profit? (e) Estimate the annual membership fee that maximizes profit. Mark this point on your graph.

Find the following: (a) \(f(g(x))\) (b) \(g(f(x))\) (c) \(f(f(x))\) $$f(x)=2 x+3\( and \)g(x)=5 x^{2}$$

You win $$\$ 38,000$$ in the state lottery to be paid in two installments -$$\$ 19,000$$ now and $$\$ 19,000$$ one year from now. A friend offers you $$\$ 36,000$$ in return for your two lottery payments. Instead of accepting your friend's offer, you take out a one-year loan at an interest rate of \(8.25 \%\) per year, compounded annually. The loan will be paid back by a single payment of $$\$ 19,000$$ (your second lottery check) at the end of the year. Which is better, your friend's offer or the loan?

You want to invest money for your child's education in a certificate of deposit (CD). You want it to be worth $$\$ 12,000$$ in 10 years. How much should you invest if the CD pays interest at a \(9 \%\) annual rate compounded (a) Annually? (b) Continuously?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.