/*! 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 51 Ice Age Many scientists believe ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 51

Ice Age Many scientists believe there have been four ice ages in the past 1 million years. Before the technique of carbon dating was known, geologists erroneously believed that the retreat of the Fourth Ice Age began about 25,000 years ago. In \(1950,\) logs from ancient spruce trees were found under glacial debris near Two Creeks, Wisconsin. Geologists determined that these trees had been crushed by the advance of ice during the Fourth Ice Age. Wood from the spruce trees contained \(27 \%\) of the level of \(^{14} \mathrm{C}\) found in living trees. Approximately how long ago did the Fourth Ice Age actually occur?

Short Answer

Expert verified
The Fourth Ice Age actually ended approximately 10,875 years ago.

Step by step solution

01

Understand Carbon Dating

Carbon dating is a method used to determine the age of ancient objects by measuring the amount of radioactive carbon-14 (^{14} Cen) left in it. Living organisms continually exchange carbon with the atmosphere through processes like respiration and photosynthesis, keeping a constant level of ^{14} C. When the organism dies, it stops exchanging carbon, and the ^{14} C starts to decay.
02

Understand Half-life of ^{14} C

The half-life of ^{14} C is approximately 5730 years. This means that every 5730 years, the amount of ^{14} C in a sample is reduced by half.
03

Determine the Fraction of ^{14} C

According to the problem, the ancient spruce trees contained 27% of the ^{14} C level found in living trees. This can be expressed as a fraction: 0.27.
04

Use the Decay Formula

The formula for radioactive decay is: N(t) = N_0 (1/2)^(t/T) where: N(t) is the remaining quantity of the substance after time t, N_0 is the initial quantity of the substance (here it's the living tree's ^{14} C level), t is the time elapsed, and T is the half-life of the substance. Since we need to find the time t, rewrite the formula to solve for t: 0.27 = (1/2)^(t/5730) .
05

Solve for t

Take the natural logarithm on both sides of the equation to solve for t: ln(0.27) = ln((1/2)^(t/5730)) . Simplify using the properties of logarithms: ln(0.27) = (t/5730) ln(1/2) .Then t = 5730 * (ln(0.27) / ln(0.5)) .Calculate t: t 鈮� 5730 * (-1.3093 / -0.6931) 鈮� 10875 years.

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 91影视!

Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process wherein unstable atomic nuclei lose energy by emitting radiation. This results in the transformation of the original nuclei into different elements or isotopes over time. In carbon dating, the focus is on the carbon-14 isotope (^{14} Cen). Living organisms maintain a constant level of ^{14} C through natural processes like breathing and feeding. When an organism dies, it stops absorbing carbon, and the existing ^{14} C begins to decay into nitrogen-14. Understanding how ^{14} C decreases in a sample lets us determine how long an organism has been dead, providing us with an age estimate.
Half-life
Half-life is the time required for half the quantity of a radioactive substance to decay. For carbon-14 (^{14} C), the half-life is approximately 5730 years. This property is crucial for determining the age of objects using carbon dating. If you start with 100% of ^{14} C in a sample, after 5730 years, only 50% remains. After another 5730 years, only 25% remains, and so forth.
A few key points to remember about half-life:
  • It's a constant value, regardless of the initial quantity of ^{14} C.
  • Allows us to create a timeline for the decay process.
  • Helps determine the time elapsed since the death of an organism by measuring the remaining ^{14} C.

In the problem mentioned, the ancient spruce trees contained 27% of the original ^{14} C, aligning this information with the half-life helps to calculate how long ago the trees had been buried.
Logarithms
Logarithms are mathematical functions used in solving equations involving exponential growth or decay, like radioactive decay.
In the context of the provided problem, the decay formula is:
N(t) = N_0 (1/2)^(t/T)
Where:
  • N(t) is the remaining quantity of the substance after time t
  • N_0 is the initial quantity of the substance
  • T is the half-life of the substance

To solve for t (the time), we use logarithms:
ln(0.27) = ln((1/2)^(t/5730))
Using logarithmic properties, it simplifies to:
ln(0.27) = (t/5730) ln(1/2)
With more simplification:
t = 5730 * (ln(0.27) / ln(0.5))
Logarithms enable us to linearize the exponential decay function, making it easier to solve for the variable t.

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

Find the present value of \(\$ 2000\) to be received in 10 years, if money may be invested at \(8 \%\) with interest compounded continuously.

Investment If real estate in a certain city appreciates at the yearly rate of \(15 \%\) compounded continuously, when will a building purchased in 2010 triple in value?

Differential Equation and Decay The amount in grams of a certain radioactive material present after \(t\) years is given by the function \(P(t) .\) Match each of the following answers with its corresponding question. Answers a. Solve \(P(t)=.5 P(0)\) for \(t.\) b. Solve \(P(t)=.5\) for \(t.\) c. \(P(.5)\) d. \(P^{\prime}(.5)\) e. \(P(0)\) f. Solve \(P^{\prime}(t)=-.5\) for \(t.\) g. \(y^{\prime}=k y\) h. \(P_{0} e^{k t}, k<0\) Questions A. Give a differential equation satisfied by \(P(t).\) B. How fast will the radioactive material be disintegrating in \(\frac{1}{2}\) year? C. Give the general form of the function \(P(t).\) D. Find the half-life of the radioactive material. E. How many grams of the material will remain after \(\frac{1}{2}\) year? F. When will the radioactive material be disintegrating at the rate of \(\frac{1}{2}\) gram per year? G. When will there be \(\frac{1}{2}\) gram remaining? H. How much radioactive material was present initially?

A news item is broadcast by mass media to a potential audience of 50,000 people. After \(t\) days, $$ f(t)=50,000\left(1-e^{-0.3 t}\right) $$ people will have heard the news. The graph of this function is shown in Fig. 8. (a) How many people will have heard the news after 10 days? (b) At what rate is the news spreading initially? (c) When will 22,500 people have heard the news? (GRAPHS CANNOT COPY) (d) Approximately when will the news spread at the rate of 2500 people per day? (e) Use equations (3) and (4) to determine the differential equation satisfied by \(f(t)\) (f) At what rate will the news spread when half the potential audience has heard the news?

The world's population was 5.51 billion on January 1, 1993, and 5.88 billion on January 1, 1998. Assume that, at any time, the population grows at a rate proportional to the population at that time. In what year will the world's population reach 7 billion?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation
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.