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 50 g of a radioactive substance and after 3 days there are only 10 g remaining, what percentage of the original amount remains after 4 days?

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,$\frac{dN}{dt}\propto N$

Given that there are 50g of a radioactive substance and after 3 days there are only 10g remain. We have to find the mass of the substance, remaining after 4 days, and its percentage of the original amount.

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.

Let the initial amount of the radioactive substance be i.e., N₀= 50g and given that the remaining amount of radioactive substance after 3 days is 10g i.e.,

t = 3 days and N = 10g

Now, from the equation (1),

$$\begin{gathered} 10=50e^{-3\lambda } \\ \frac{1}{5}=e^{-3\lambda } \\ {e}^{3\lambda }=5 \\ 3\lambda =ln5 \\ \lambda =\frac{ln5}{3} \\ \lambda =0.5365 \end{gathered}$$

One will use this value of$\lambda$ in the next step to find the value of the mass of the remaining radioactive substance after 4days.

Now we will find the mass of the remaining radioactive substance after 4 days.

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

N₀=50 g

Time, t = 4 days

(From Step 3)

Using the equation (1),

$$\begin{gathered} N=N_0{e}^{-\lambda t} \\ N=\left(50\right)·e^{-\left(0.5365\right)4} \\ N=5.847g \end{gathered}$$

Hence, the mass of the remaining radioactive substance after 4 days is 5.847 g.

The mass of the remaining radioactive substance after 4 days is 5.847 g

Therefore,

Percentage of remaining mass

$$\begin{gathered} =\frac{5.847}{N_0}\times 100 \\ =\frac{5.847}{50}\times 100 \\ =11.7\% \end{gathered}$$

Thus, the percentage of the original amount that remains after 4 days is 11.7%.