# Iodine-131 is a radioactive isotope with a half-life of 8 days. How many grams of a 64 g sample of iodine-131 will remain at the end of 24 days?

May 25, 2016

$\text{8 g}$

#### Explanation:

The nuclear half-life of a radioactive isotope tells you how much time must pass in order for half of the atoms present in an initial sample to undergo radioactive decay.

In essence, the half-life tells you at what time intervals you can expect an initial sample of a radioactive isotope to be halved.

In your case, iodine-131 is said to have a half-life of $8$ days. This means that with every $8$ days that pass, your sample of iodine-131 will be halved.

If you take ${A}_{0}$ to be the initial sample of iodine-131, you will be left with

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

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

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

and so on.

You can thus say that the amount of iodine-131 that remains undecayed, $\text{A}$, after a period of time $t$, will be equal to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} A = {A}_{0} \cdot \frac{1}{2} ^ n \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here $n$ represents the number of half-lives that pass in $t$.

$n = \left(24 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{days"))))/(8color(red)(cancel(color(black)("days}}}}\right) = 3$

This will get you

${A}_{\text{24 days" = "64 g}} \cdot \frac{1}{2} ^ 3$

A_"24 days" = color(green)(|bar(ul(color(white)(a/a)"8 g"color(white)(a/a)|)))

May 25, 2016

$\implies {m}_{t} = 8 g$

#### Explanation:

Half Life follows first Order kinetics

This graph shows what is going on during the decay

As you can see from the graph the half life is 8 days

Lets do the math and verify it with the graph

Recall

${t}_{\frac{1}{2}} = \ln \frac{2}{k}$

$8 = \ln \frac{2}{k}$

$k = \ln \frac{2}{8}$

With this you can solve most problems regarding the above data

$\ln \left({A}_{o} / {A}_{t}\right) = k t$

$\ln \left({A}_{o} / {A}_{t}\right) = \ln \frac{2}{8} \cdot 24 = 3 \ln 2 = \ln {2}^{3} = \ln 8$

$\left({A}_{o} / {A}_{t}\right) = 8$

Now

Greater the concentration of a sample greater the mass in in it

$C \propto M$

So

$\left({A}_{o} / {A}_{t}\right) = \left({m}_{o} / {m}_{t}\right) = 8$

$\text{we have "m_o = 64 g}$

$\frac{64}{m} _ t = 8$

$\implies {m}_{t} = 8 g$

Verifying with the graph

m_("24 hours") = 12.5 % m_o = 12.5 *10^-2 *m_o
$= 12.5 \cdot {10}^{-} 2 \cdot 64 = 8 g$

Hence verified