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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.

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2 Answers 2

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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 $x=\sqrt2g$.

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  • $\begingroup$ How is the disintegration from day $4$ to day $5$ obtained ?How do we know that $\cfrac{2-x}{x}$ is the amount of material in the beginning of the $6^{th} $ day ?This is the only thing I can't understand. $\endgroup$
    – Mr. Y
    Commented Jan 12, 2016 at 10:43
  • $\begingroup$ You are confusing the quantities and the decay. The daily decay is a constant, computed as (initial amount-final amount)/initial amount. The quantities are as listed per day. $\endgroup$
    – user65203
    Commented Jan 12, 2016 at 10:46
  • $\begingroup$ What is the intuition geometrically ?Is the decay the slope between these points ? $\endgroup$
    – Mr. Y
    Commented Jan 12, 2016 at 10:55
  • $\begingroup$ No, the slope would be $\Delta q$ (for $\Delta t=1$). The decay is $\Delta q/q$, it is a "relative slope". $\endgroup$
    – user65203
    Commented Jan 12, 2016 at 10:57
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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:

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

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  • $\begingroup$ This is even harder to understand.Thanks anyway. $\endgroup$
    – Mr. Y
    Commented Jan 12, 2016 at 10:24
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    $\begingroup$ @Mr.Y What's too hard to understand? I can elaborate further. The only thing I used is a simple exponential function. If you don't know what those are, you should study up on those before solving problems about half life. $\endgroup$
    – 5xum
    Commented Jan 12, 2016 at 10:27
  • $\begingroup$ Problem is that the formula you've used is taught after some exercises on my book,so there's some simpler argument to solve the problem before using the derived formula for half life decay. $\endgroup$
    – Mr. Y
    Commented Jan 12, 2016 at 10:30
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    $\begingroup$ @Mr.Y And at this point an inconvenient feature of your question hits us in the face, which is that we do not know what you know and what you do not know. In such a state of ignorance, 5xum's answer is a worthy try. If all you know is that after a multiple $k$ of the halflife the remaining quantity is the initial one divided by $2^k$, there is no way you can solve this exercise... $\endgroup$
    – Did
    Commented Jan 12, 2016 at 10:36
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    $\begingroup$ @did: anyway, the OP did explain that the result is obtained from a proportionality relation (which amounts to a quadratic equation) and his question concerns the latter. $\endgroup$
    – user65203
    Commented Jan 12, 2016 at 11:05

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