Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 69
Suppose that the rate of change of the mass \(m\) of a sample of the isotope \({ }^{14} C\) satisfies $$m^{\prime}(t)=-0.1213 \cdot e^{-0.0001213 t} \frac{\mathrm{g}}{\mathrm{yr}}$$ when \(t\) is measured in years. If \(m(0)=1000 \mathrm{~g},\) then for what value of \(t\) is \(m(t)\) equal to \(800 \mathrm{~g}\) ?
Short Answer
Expert verified
It takes approximately 1579.45 years for the mass to reduce to 800 g.
Step by step solution
01
Understand the Problem
We have a differential equation describing how the mass of a radioactive isotope changes over time: \( m'(t) = -0.1213 \cdot e^{-0.0001213t} \frac{\mathrm{g}}{\mathrm{yr}} \). We want to find the time \( t \) when the mass \( m(t) \) equals \( 800 \mathrm{~g} \) given the initial condition \( m(0) = 1000 \mathrm{~g} \).
02
Set Up the Integral for Solution
We aim to integrate the rate of change \( m'(t) \) to find \( m(t) \). The integral we need to solve is:\[m(t) = \int -0.1213 \cdot e^{-0.0001213t} \, dt + C\]Where \( C \) is the constant of integration.
03
Integrate the Differential Equation
The integration of \(-0.1213 \cdot e^{-0.0001213t} \) gives:\[m(t) = 1000 - e^{-0.0001213t} + C\]In this case, the correct integration of this specific case results in:\[m(t) = -10^3 e^{-0.0001213t} + C\] We then apply the initial condition \( m(0) = 1000 \) to find \( C \).
04
Apply Initial Condition to Determine C
Substitute \( m(0) = 1000 \) into the integrated function:\[1000 = -10^3 e^{0} + C\]This simplifies to:\[1000 = -10^3 + C\]Solving for \( C \), we find \( C = 2000 \). The equation is then:\[m(t) = -10^3 e^{-0.0001213t} + 2000\]
05
Solve for t When m(t) = 800
Set \( m(t) = 800 \) and solve for \( t \):\[800 = -10^3 e^{-0.0001213t} + 2000\]Rearranging gives:\[-10^3 e^{-0.0001213t} = 800 - 2000 = -1200\]Thus, \( e^{-0.0001213t} = \frac{1200}{10^3} = 1.2 \).Take the natural logarithm to solve for \( t \):\[-0.0001213t = \ln(1.2)\]\[t = \frac{\ln(1.2)}{-0.0001213}\]
06
Calculate the Value of t
Compute \( t \) using the logarithm:\[t = \frac{\ln(1.2)}{-0.0001213} \approx 1579.45\]Thus, the value of \( t \) for which \( m(t) = 800 \mathrm{~g} \) is approximately 1579.45 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 by which an unstable atomic nucleus loses energy by emitting radiation. In this context, we are working with the decay of radioactive isotopes over time. This decay is usually exponential, meaning the substance decreases by a consistent percentage over time.
It is important to remember:
It is important to remember:
- Radioactive decay is random on the atomic scale but predictable on a larger scale.
- The decay rate is often expressed through a half-life, which is the time needed for half of the radioactive nuclei to decay.
- The rate of decay depends on the stability of the isotope and is independent of external conditions like temperature and pressure.
Integration
Integration is a fundamental concept in calculus used to find quantities like area under a curve or total change over an interval. In this exercise, integration allows us to determine the mass function \( m(t) \) from its rate of change \( m'(t) \).
The integral to find \( m(t) \) is:
The integral to find \( m(t) \) is:
- \[ m(t) = \int -0.1213 \cdot e^{-0.0001213t} \, dt + C \]
- First, recognize the integral of an exponential function. Use the formula: \( \int e^{ax} \, dx = \frac{1}{a}e^{ax} + C \).
- Apply the constants in the problem and solve: \( \int -0.1213 \cdot e^{-0.0001213t} \, dt \) yields \(-10^3 e^{-0.0001213t} + C \).
Initial Value Problem
An initial value problem (IVP) involves finding a function based on its rate of change and an initial condition. In this exercise, the initial value problem is presented with:\( m(0) = 1000 \text{g} \)This means that at time \( t=0 \), the mass of the radioactive substance is \( 1000 \text{g} \).
Using this information, we solve for the constant of integration \( C \) in the function \( m(t) \):
Using this information, we solve for the constant of integration \( C \) in the function \( m(t) \):
- After integrating, we have \[ m(t) = -10^3 e^{-0.0001213t} + C \].
- Substitute \( t = 0 \) and \( m(0) = 1000 \) to find \( C \):
- \( 1000 = -10^3 \cdot e^{0} + C \), simplify to \( 1000 = -1000 + C \)
- This results in \( C = 2000 \).
Exponential Functions
Exponential functions form the backbone of this radioactive decay problem. An exponential function is a mathematical function of the form \( f(x) = a \cdot e^{bx} \), where \( e \) is the base of natural logarithms, approximately equal to 2.71828.
Key attributes of exponential functions in the context of this problem include:
\[ m(t) = -10^3 e^{-0.0001213t} + 2000\] describes an exponential decay model, where \(-0.0001213\) signifies a decay rate. To find when \( m(t) = 800 \text{g} \), we solve:
Key attributes of exponential functions in the context of this problem include:
- They model processes that change multiplicatively, such as population growth or radioactive decay.
- The base \( e \) ensures continuous growth or decay.
- Exponential decay occurs when the exponent \( b \) is negative, leading to a decreasing outcome over time.
\[ m(t) = -10^3 e^{-0.0001213t} + 2000\] describes an exponential decay model, where \(-0.0001213\) signifies a decay rate. To find when \( m(t) = 800 \text{g} \), we solve:
- \[800 = -10^3 e^{-0.0001213t} + 2000\]
- Rearrange and isolate the exponential term to calculate \( t \).
- The expression \( e^{-0.0001213t} = 1.2 \) comes from simplifying the mass equation step-by-step.
- Taking the natural logarithm, solve for \( t \):
\[ t = \frac{\ln(1.2)}{-0.0001213} \approx 1579.45 \text{ years} \]
