How much of radioactive material substance will remain after $5$ days?

Question

Initially there are $8$ grams of a radioactive material in a container.The half-life of the material is $2$ days.

How much of the radioactive substance will remain after $5$ days ?

By exponential decay I know that after $2$ days I have $4$ grams of the radiactive material and that $2$ days later I will have $\cfrac{2}{2}=1$ gram of the substance.

Now I don't know how to find how much radioactive I have after $5$ days ,i.e. in the "beginning",roughly speaking, of the $6^{th}$ day.

My book provides as a solution the following equation $\cfrac{2-x}{2}=\cfrac{x-1}{x}$ ($x$ is the amount of material after $5$ days) from which it follows that $x=\sqrt{2}$ but I don't know how this equation have been set up,what is meaning behind this.

If someone can help me understand how that equation is derived I am really grateful.

Answer 1

The remaining quantities at the beginning of the days are known to be

$$0\to8g\\2\to4g\\4\to2g\\5\to xg\\6\to1g.$$

Then you express that the relative quantity that disappears in a single day (the daily decay) is a constant. With the disintegrations from day $4$ to day $5$ and day $5$ to day $6$,

$$\frac{2-x}2=\text{Cst}=\frac{x-1}{x}.$$

It is simpler to express the fraction that remains, giving $$\frac x2=\text{Cst}=\frac 1x$$ and you immediately get $x=\sqrt2g$.

Answer 2

The amount of material after time $t$ (let's forget units for now) is $$m(t) = m(0) \cdot 2^{-\frac t{\tau}}$$

where $\tau$ is the half-life of the material (you can see that plugging in $t=\tau$ gives you $m(\tau)=m(0)\cdot\frac12$ which is what you would expect).

Now, what you want to know is what $m(t)$ will be equal to when $t=5$, given that $\tau=2$ and $m(0)=8$. Simply plug in the numbers and voila:

$$m(5)=8g\cdot 2^{-\frac{5\text{ days}}{2\text{ days}}} = \frac{8g}{2^\frac{5}{2}} = \frac{8}{2\cdot2\cdot\sqrt2}g = \sqrt2g$$