Q 3.2-24E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 23–27, assume that the rate of decay of a radioactive substance is proportional to the amount of the substance present. The half-life of a radioactive substance is the time it takes for one-half of the substance to disintegrate. If initially there are 300 g of a radioactive substance and after 5 yr there are 200 g remaining, how much time must elapse before only 10 g remain?

Step-by-Step Solution

Answer

10g of the radioactive substance will remain after 41.9 years. 

Step1: Analyzing the given statement

Given that the rate of decay of a radioactive substance is directly proportional to the amount of the substance present.  Let the present amount of the radioactive substance be N.

Therefore, $\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{\propto }\mathit{N}$

Given that there are 300g of a radioactive substance and after 5 years there are only 200g remaining. We have to find the time elapsed, before only 10g remain.   

Step 2: Determining the formula with the help of the given proportionality relation, to solve the question

Given, 

$$\begin{gathered} \frac{dN}{dt}\propto N \\ \frac{dN}{dt}=-\lambda N \end{gathered}$$

 where, $\lambda$ is the constant of proportionality.

 $$\begin{gathered} \frac{dN}{N}=-\lambda \\ \int \frac{dN}{N}=-\lambda \int dt \\ lnN=-\lambda t+ln{N}_0 \end{gathered}$$

where, In N₀ is an arbitrary constant.

 $$\begin{gathered} lnN-ln{N}_0=-\lambda t \\ ln\left(\frac{N}{N_0}\right)=-\lambda t \\ \frac{N}{N_0}=e^{-\lambda t} \\ N=N_0{e}^{-\lambda t}······\left(1\right) \end{gathered}$$

One will use this formula to solve the question.

 

Step 3: Using the formula obtained in the step 2, we will find the value of λ

Let the initial amount of the radioactive substance be N₀ i.e.,N₀ = 300g and given that the remaining amount of radioactive substance after 5 years is 200g i.e., 

 t = 5 years and N = 200g

 Now, from the equation (1),

 $$\begin{gathered} 200=300e^{-5\lambda } \\ \frac{2}{3}=e^{-5\lambda } \\ {e}^{5\lambda }=1.5 \\ 5\lambda =ln\left(1.5\right) \\ \lambda =\frac{ln\left(1.5\right)}{5} \\ \lambda =0.0811 \end{gathered}$$

One will use this value of $\lambda$ in the next step to finding the time elapsed, before only 10g of the substance remain.   

 

Step 4: Finding the time elapsed, before only 10g of the substance remain

For this, let N be the mass to be found,  

N₀=300 g

N = 10 g

(From Step 3)

 Using the equation (1),

$$\begin{gathered} N=N_0{e}^{-\lambda t} \\ 10=\left(300\right)·e^{-\left(0.0811\right)t} \\ \frac{1}{30}=e^{-\left(0.0811\right)t} \\ e^{\left(0.0811\right)t}=30 \\ \left(0.0811\right)t=ln30 \\ t=\frac{ln30}{0.0811} \\ t=41.9 years \end{gathered}$$

Hence, 10g of a radioactive substance will remain after 41.9 years.