Question Statement

Most radioactive substances disintegrate at a rate proportional to the amount present. If the initial amount of a radioactive substance is $50$ grams and its half-life is $1000$ years, find the amount of substance present after $800$ years.

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Background and Explanation

This problem is modeled using a first-order linear differential equation representing exponential decay. The rate of change of the quantity is proportional to the quantity itself, leading to the standard decay formula $A(t)=A_0{e}^{-kt}$.

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Solution

Let $A$ be the amount of radioactive substance present at any time $t$.

Step 1: Set up the Differential Equation

According to the problem, the rate of disintegration is proportional to the amount present:

$\frac{dA}{dt}\propto A$

To turn this into an equation, we introduce a decay constant $k$:

$\frac{dA}{dt}=-kA$

(Note: The negative sign represents the decay or decrease of the substance over time.)

Step 2: Solve the Differential Equation

We use the method of separation of variables:

$\frac{dA}{A}=-kdt$

Integrating both sides:

$\int \frac{dA}{A}=-k\int dt$

$\ln A=-kt+c$

Converting from logarithmic to exponential form:

$A=e^{-kt+c}$

Step 3: Find the Constant of Integration

We are given that at $t=0$, the initial amount $A=50$ grams. Substituting these values into equation (1):

$\begin{array}{rl}50& =e^{-k(0)+c}\\ 50& =e^c\\ \ln 50& =c\end{array}$

Substitute $c$ back into equation (1):

$\begin{array}{rl}A& =e^{-kt+\ln 50}\\ A& =e^{-kt}\cdot e^{\ln 50}\\ A& =50e^{-kt}\end{array}$

Step 4: Determine the Decay Constant ($k$)

The half-life is $1000$ years, meaning at $t=1000$, the amount $A$ becomes half of the initial amount ($A=\frac{50}{2}=25$). Substituting this into equation (2):

$\begin{array}{rl}25& =50e^{-k(1000)}\\ \frac{25}{50}& =e^{-1000k}\\ \frac{1}{2}& =e^{-1000k}\end{array}$

Taking the natural logarithm ($\ln$) of both sides:

$\begin{array}{rl}\ln \left(\frac{1}{2}\right)& =\ln e^{-1000k}\\ \ln 1-\ln 2& =-1000k\cdot \ln e\\ 0-\ln 2& =-1000k\cdot (1)\\ \ln 2& =1000k\\ k& =\frac{\ln 2}{1000}\end{array}$

Substituting $k$ back into equation (2) gives the general model for this substance:

$A=50e^{-\left(\frac{\ln 2}{1000}\right)t}$

Step 5: Calculate the Amount after 800 Years

To find the amount present after $t=800$ years, substitute $t=800$ into equation (3):

$A=50\cdot e^{-\left(\frac{\ln 2}{1000}\right)(800)}$

Simplify the exponent:

$\begin{array}{rl}A& =50\cdot e^{-\frac{800}{1000}\ln 2}\\ A& =50\cdot e^{-0.8\ln 2}\end{array}$

Using the values $\ln 2\approx 0.6931$:

$\begin{array}{rl}A& =50\cdot e^{-(0.8)(0.6931)}\\ A& =50\cdot e^{-0.5545}\\ A& =50\cdot (0.5744)\\ A& \approx 28.72\text{ grams}\end{array}$

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Key Formulas or Methods Used

• Law of Radioactive Decay: $\frac{dA}{dt}=-kA$
• Separation of Variables: Method for solving first-order differential equations.
• Exponential Growth/Decay Formula: $A=A_0{e}^{-kt}$
• Half-life Relation: $k=\frac{\ln 2}{T_{1/2}}$

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Summary of Steps

1. Model the problem using the differential equation $\frac{dA}{dt}=-kA$.
2. Solve the equation to find the general form $A=e^{-kt+c}$.
3. Apply initial conditions ($t=0,A=50$) to find the specific equation $A=50e^{-kt}$.
4. Use the half-life ($t=1000,A=25$) to solve for the decay constant $k$.
5. Substitute the target time ($t=800$) into the final equation to calculate the remaining mass.