/*! 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 6 The mass of a radioactive sample... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The mass of a radioactive sample decays at a rate that is proportional to its mass. a. Express this fact as a differential equation for the mass \(M(t)\) using \(k\) for the constant of proportionality. b. If the initial mass is \(M_{0}\), find an expression for the mass \(M(t)\). c. The half-life of the sample is the amount of time required for half of the mass to decay. Knowing that the half-life of Carbon-14 is 5730 years, find the value of \(k\) for a sample of Carbon-14. d. How long does it take for a sample of Carbon-14 to be reduced to one- quarter its original mass? e. Carbon-14 naturally occurs in our environment; any living organism takes in Carbon14 when it eats and breathes. Upon dying, however, the organism no longer takes in Carbon-14. Suppose that you find remnants of a pre-historic firepit. By analyzing the charred wood in the pit, you determine that the amount of Carbon-14 is only \(30 \%\) of the amount in living trees. Estimate the age of the firepit.

Short Answer

Expert verified
The differential equation is \( \frac{dM}{dt} = -kM \). The expression for mass is \( M(t) = M_{0}e^{-kt} \). The value of \( k \) is approximately \( 1.2104 \times 10^{-4} \text{years}^{-1} \). It takes about 11460 years to decay to one-quarter. The firepit is approximately 10000 years old.

Step by step solution

01

- Set up the differential equation

The mass of the radioactive sample decays at a rate proportional to its mass. This can be mathematically expressed using the differential equation \[ \frac{dM}{dt} = -kM \] where \( k \) is the constant of proportionality.
02

- Solve the differential equation

To solve the differential equation \( \frac{dM}{dt} = -kM \), separate the variables and integrate:\[ \int \frac{1}{M} dM = -k \int dt \]This gives \[ \ln M = -kt + C \]Exponentiating both sides, we find \[ M(t) = M_{0}e^{-kt} \]where \( M_{0} \) is the initial mass.
03

- Find the decay constant \( k \)

The half-life \( T_{1/2} \) of Carbon-14 is 5730 years. Using the relation \[ M(5730) = \frac{1}{2}M_{0} \], we can write \[ \frac{M_{0}}{2} = M_{0}e^{-k \cdot 5730} \].Dividing both sides by \( M_{0} \) and simplifying, we get \[ \frac{1}{2} = e^{-k \cdot 5730} \], Taking the natural logarithm of both sides, we find \[ -5730k = \text{ln}\frac{1}{2} \] or \[ k = \frac{-\text{ln}(1/2)}{5730} \] \[ k = \frac{\text{ln} 2}{5730} \] \[ k \approx 1.2104 \times 10^{-4} \text{years}^{-1} \].
04

- Time to decay to one-quarter

To find the time \( t \) when the sample is reduced to one-quarter of its original mass, we use \[ M(t) = \frac{M_{0}}{4} = M_{0}e^{-kt}. \] Dividing both sides by \( M_{0} \) and simplifying, we get \[ \frac{1}{4} = e^{-kt} \], Taking the natural logarithm of both sides, we find \[ -kt = \text{ln}\frac{1}{4} \], \[ t = \frac{-\text{ln}(1/4)}{k} \], \[ t = \frac{\text{ln} 4}{1.2104 \times 10^{-4}} \], \[ t \approx 11460 \text{ years} \].
05

- Age of the firepit

If the amount of Carbon-14 is only 30% of the original amount, we use \[ M(t) = 0.3M_{0} = M_{0}e^{-kt}. \] Dividing both sides by \( M_{0} \) and simplifying, we get \[ 0.3 = e^{-kt} \], Taking the natural logarithm of both sides, we find \[ -kt = \text{ln}0.3 \], \[ t = \frac{\text{ln}(1/0.3)}{k} \], \[ t \approx \frac{1.204}{1.2104 \times 10^{-4}} \approx 10000 \text{ years} \].

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 91Ó°ÊÓ!

Key Concepts

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

differential equations
Differential equations are mathematical equations that relate a function with its derivatives. They are used to model various real-world phenomena by describing how a system changes over time.
In the context of radioactive decay, we use a differential equation to represent how the mass of a radioactive sample decreases over time. The key idea is that the rate of decay is proportional to the current mass.
For a mass \(M(t)\), we can express this with the differential equation: \ \frac{dM}{dt} = -kM \ Here, \ \frac{dM}{dt} \ represents the rate of change or decay of the mass \(M(t)\),
and \(k\) is the constant of proportionality that defines the decay rate.
radioactive decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This process transforms the original nucleus into a different state, often into a different element.
The decay process follows an exponential pattern, depicted by the equation: \ M(t) = M_{0}e^{-kt} \ Here, \(M(t)\) is the mass of the radioactive sample at time \(t\), and \(M_{0}\) is the initial mass of the sample.
The term \(e^{-kt}\) indicates that the mass decreases exponentially over time, with \(k\) being the decay constant. This formula is derived by solving the differential equation \ \frac{dM}{dt} = -kM \ which expresses that the rate of decay is proportional to the current mass.
half-life calculations
The half-life of a radioactive substance is the time required for half of the sample to decay. It is a unique characteristic of each radioactive isotope.
We can use the half-life to find the decay constant \(k\). For example, the half-life of Carbon-14 is 5730 years. The formula to calculate \(k\) is: \ k = \frac{\text{ln}(2)}{T_{1/2}} \ Here, \(T_{1/2}\) is the half-life. Plugging in the half-life of Carbon-14, we have: \ k = \frac{\text{ln}(2)}{5730} \ This value of \(k\) helps in determining how the mass decays over time.
Carbon-14 dating
Carbon-14 dating, also known as radiocarbon dating, is a method for determining the age of an object containing organic material. This technique utilizes the known half-life of Carbon-14, which is 5730 years.
When an organism dies, it no longer absorbs Carbon-14, and the Carbon-14 it contains starts to decay at a predictable rate. By measuring the remaining amount of Carbon-14, we can estimate the age of the sample.
For instance, if a sample contains only 30% of its original Carbon-14, we can set up the equation: \ 0.3M_{0} = M_{0}e^{-kt} \ Solving for \(t\), we determine the sample's approximate age, in this case, nearly 10,000 years. This technique is invaluable in archaeology and other fields studying ancient organic materials.

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

Newton's Law of Cooling says that the rate at which an object, such as a cup of coffee, cools is proportional to the difference in the object's temperature and room temperature. If \(T(t)\) is the object's temperature and \(T_{r}\) is room temperature, this law is expressed at $$ \frac{d T}{d t}=-k\left(T-T_{r}\right) $$ where \(k\) is a constant of proportionality. In this problem, temperature is measured in degrees Fahrenheit and time in minutes. a. Two calculus students, Alice and Bob, enter a \(70^{\circ}\) classroom at the same time. Each has a cup of coffee that is \(100^{\circ} .\) The differential equation for Alice has a constant of proportionality \(k=0.5,\) while the constant of proportionality for Bob is \(k=0.1 .\) What is the initial rate of change for Alice's coffee? What is the initial rate of change for Bob's coffee? b. What feature of Alice's and Bob's cups of coffee could explain this difference? c. As the heating unit turns on and off in the room, the temperature in the room is $$ T_{r}=70+10 \sin t $$ Implement Euler's method with a step size of \(\Delta t=0.1\) to approximate the temperature of Alice's coffee over the time interval \(0 \leq t \leq 50 .\) This will most easily be performed using a spreadsheet such as Excel. Graph the temperature of her coffee and room temperature over this interval. d. In the same way, implement Euler's method to approximate the temperature of Bob's coffee over the same time interval. Graph the temperature of his coffee and room temperature over the interval. e. Explain the similarities and differences that you see in the behavior of Alice's and Bob's cups of coffee.

The Gompertz equation is a model that is used to describe the growth of certain populations. Suppose that \(P(t)\) is the population of some organism and that $$ \frac{d P}{d t}=-P \ln \left(\frac{P}{3}\right)=-P(\ln P-\ln 3) $$ a. Sketch a slope field for \(P(t)\) over the range \(0 \leq P \leq 6\). b. Identify any equilibrium solutions and determine whether they are stable or unstable. c. Find the population \(P(t)\) assuming that \(P(0)=1\) and sketch its graph. What happens to \(P(t)\) after a very long time? d. Find the population \(P(t)\) assuming that \(P(0)=6\) and sketch its graph. What happens to \(P(t)\) after a very long time? e. Verify that the long-term behavior of your solutions agrees with what you predicted by looking at the slope field.

Consider the initial value problem $$ \frac{d y}{d t}=-\frac{t}{y}, y(0)=8 $$ a. Find the solution of the initial value problem and sketch its graph. b. For what values of \(t\) is the solution defined? c. What is the value of \(y\) at the last time that the solution is defined? d. By looking at the differential equation, explain why we should not expect to find solutions with the value of \(y\) you noted in (c).

Find the solution to the differential equation $$ \begin{array}{l} \frac{d y}{d t}=y^{2}(5+t), \\ y=5 \text { when } t=1 . \end{array} $$

Solve the separable differential equation for \(u\) $$ \frac{d u}{d t}=e^{2 u+8 t} $$ Use the following initial condition: \(u(0)=13\). \(=\)

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.