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?

Short answer

10g of the radioactive substance will remain after41.9 years.

Step-by-step solution

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.

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.

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.

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 = 10g

(From Step 3)

Using the equation (1),

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

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