Question

A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is $50$ kg of the material present and after two hours it is observed that the material has lost $10$ percent of its original mass. The expression for the mass of the material remaining at any time $t$ is:

• A. $N=50e^{\left(\frac{1}{2}\right)\left(\ln 0.9\right)t}$
• B. $N=50e^{-\left(\frac{1}{4}\right)\left(\ln 9\right)t}$
• C. $N=50e^{-\left(\ln 0.9\right)t}$
• D. $N=50e^{-\left(\frac{1}{2}\right)\left(\ln 0.9\right)t}$

Answer 1

The correct option is A $N=50e^{\left(\frac{1}{2}\right)\left(\ln 0.9\right)t}$

Let $N$ denotes the amount of material present at time $t$.
Then, $\frac{dN}{dt}=-kN$
This differential equation is separable and linear, its solution is $N=c{e}^{-kt}\cdots \left(1\right)$
At $t=0,$ we are given that $N=50.$ Therefore from the equation $\left(1\right)$ we have $50=c{e}^{k\left(0\right)},\Rightarrow c=50$
Thus, $N=50e^{-kt}\cdots \left(2\right)$
At $t=2,10$ percent of the original mass of $50$ kg or $5$ kg, has decayed.
Hence, at $t=2,N=45.$
Substituting these values into equation $\left(2\right)$ and solving for $k$,
we have $45=50e^{-2k}\Rightarrow k=\frac{1}{2}\ln \frac{50}{45}$
Substituting these values into equation $\left(2\right)$, we obtain the amount of mass present at any time $t$ as
$N=50e^{\left(\frac{1}{2}\right)\left(\ln 0.9\right)t}\cdots \left(3\right)$
where $t$ is measured in hours.