# Question bb66f

Jun 15, 2016

Here's what I got.

#### Explanation:

A radioactive nuclide's nuclear half-life is defined as the time needed for half of an initial sample to undergo radioactive decay.

If you take ${\text{N}}_{0}$ to be the initial concentration of your radioactive nuclide $\text{X}$, you can say that you will be left with

$\frac{1}{2} \cdot {\text{N"_0 = "N}}_{0} / 2 \to$ after the passing of one half-life

$\frac{1}{2} \cdot {\text{N"_0/2 = "N}}_{0} / 4 \to$ after the passing of two half-lives

$\frac{1}{2} \cdot {\text{N"_0/4 = "N}}_{0} / 8 \to$ after the passing of three half-lives
$\vdots$

and so on.

This means that you can express the amount of the nuclide that remains undecayed after a period of time $t$ like this

color(blue)(|bar(ul(color(white)(a/a)"N"_t = "N"_0 * 1/2^ncolor(white)(a/a)|)))" " " "color(orange)("(*)")

Here

$n$ - the number of half-lives that pass in the given period of time $t$

All you have to do now is pick a period of time $t$ and use the known ratio that exists between ${\text{N}}_{t}$ and ${\text{N}}_{0}$ to find $n$.

For example, for $t = 0 \to t = \text{5 h}$ you have

${\text{N"_t/"N}}_{0} = \frac{1}{2} ^ n = 0.6484$

This is equivalent to

${2}^{n} = \frac{1}{0.6484}$

$\ln \left({2}^{n}\right) = \ln \left(\frac{1}{0.6484}\right)$

$n \cdot \ln \left(2\right) = \ln \left(\frac{1}{0.6484}\right) \implies n = \ln \frac{\frac{1}{0.6484}}{\ln} \left(2\right) = 0.625$

So, you know that $0.625$ half-lives pass in $5$ hours, which means that the half-life of the nuclide, ${t}_{\text{1/2}}$, is

t_"1/2" = "5 h"/0.625 = color(green)(|bar(ul(color(white)(a/a)color(black)("8 h")color(white)(a/a)|)))

I recommend using different time periods to find $n$, the half-life must come out the same in every case. That is the case because radioactive decay is a first-order reaction, meaning that its half-life is constant throughout the decay process.

For example, for $t = 0 \to t = \text{18 h}$ you have

${\text{N"_t/"N}}_{0} = 0.2102$

This time you get

$n = \ln \frac{\frac{1}{0.2102}}{\ln} \left(2\right) = 2.25$

Once again, you have

${t}_{\text{1/2" = "18 h"/2.25 = "8 h}}$

For the second part, use the half-life of the reaction to find the value of $n$

$n = \left(64 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{h"))))/(8color(red)(cancel(color(black)("h}}}}\right) = 8$

then use equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to find the mass of the nuclide that remains undecayed

$\text{N"_t = "2.500 mg} \cdot \frac{1}{2} ^ 8$

$\text{N"_t = "0.009766 mg}$

This means that the amount of the nuclide that decayed is

"N"_"decayed" = "2.500 mg" - "0.009766 mg" = color(green)(|bar(ul(color(white)(a/a)color(black)("2.490 mg")color(white)(a/a)|)))#

I'll leave the answer rounded to four sig figs.

Oct 20, 2016

The half life is 8 hours.

#### Explanation:

O.K., this is actually one that requires almost no math to solve!

As you know, or are learning, a half-life is the amount of time it takes for half of a radioactive element to decay. So, in the language of your question Nt/N0 = 0.5 at one half life.

Looking at your chart, we see that that happened somewhere between 5 and 10 hours.

Now we also know that after two half lives Nt/N0 will be 0.25.

Woa!!!! THAT's ON THE CHART!!!! Awesome!

Two half-lives = 16 hours.. Therefore one half-life = 8 hours.

Wait! Does that check with the previous observation that the half-life will be between 5 and 10 hours? Yup! O.K.! We are in business!

On to part B! 64 hours is 64/8 half-lives = 8 half-lives. After 8 half-lives there will be $\frac{1}{2} ^ 8$ of the original substance remaining. or 0.00390625 * original mass remaining. So we have lost (2.500 - (0.00390625*2.500)) grams, or 2.4902 grams of the original material.