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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,$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{鈭�}\mathit{N}$

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.

Given,

$$\begin{gathered} \frac{dN}{dt}鈭�N \\ \frac{dN}{dt}=-位N \end{gathered}$$

where, $位$is the constant of proportionality.

$$\begin{gathered} \frac{dN}{N}=-位 \\ 鈭�\frac{dN}{N}=-位鈭�dt \\ lnN=-位t+ln{N}_0 \end{gathered}$$

where, In N₀ is an arbitrary constant.

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

One will use this formula to solve the question.

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),

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

One will use this value of $位$in the next step to 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),

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

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