# Question e121a

Jul 9, 2015

You will be left with $\frac{1}{128}$ of the original sample.

#### Explanation:

A radioactive isotope's nuclear half-life tells you how much time must pass before the half of the original sample undergoes decay.

In your case, the half-life of iodine-131 is 8 days. This means that any sample of iodine-131 that undergoes decay will be halved after 8 days, halved again after another 8 days, halved again after another 8 days, and so on.

You know that the fuel rods are stored underwater for 56 days. You can use this value to determine how many half-lives will pass.

$\text{no. of half-lives" = n_"half" = t/t_"1/2}$, where

${t}_{\text{1/2}}$ - the half-life of the element.

Plug in your values to get

n_"half" = (56cancel("days"))/(8cancel("days")) = 7#

So, if the initial amount gets halved with every passing half-life, you can write

$A \left(t\right) = {A}_{0} / {2}^{{n}_{\text{half}}}$

$A \left(t\right) = {A}_{0} / {2}^{7} = \frac{1}{128} \cdot {A}_{0}$

This means that the fraction remaining after 56 days will be

$A \frac{t}{A} _ 0 = \frac{\frac{1}{128} \cdot \cancel{A \left[0\right]}}{\cancel{A \left[0\right]}} = \textcolor{g r e e n}{\frac{1}{128}}$

Jul 9, 2015

1/128 remains

#### Explanation:

The first thing to do is count how many half lives have elapsed.

The half life (${t}_{\frac{1}{2}}$ )is the time taken for half the amount of a given sample to decay.

You can see that 56 days is 7 half lives since 56/8 = 7

To get the fraction remaining we need to x by 1/2 seven times:

1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/128

If you don't get nice numbers like this you can use the equation for radioactive decay:

${N}_{t} = {N}_{0} {e}^{- \lambda t}$

${N}_{t}$ = no. atoms remaining

${N}_{0}$ = initial no. atoms

$t$ = time elapsed

$\lambda$ = the decay constant.

We get the decay constant from the half life:

$\lambda = \frac{0.693}{{t}_{\frac{1}{2}}}$

$\lambda = \frac{0.693}{8} = 0.0866 {d}^{- 1}$

Taking natural logs$\Rightarrow$

$\ln \left(\frac{{N}_{t}}{{N}_{0}}\right) = - \lambda \times t = - 0.0866 \times 56 = - 4.851$

From which $\frac{{N}_{t}}{{N}_{0}} = 0.00782$

To compare this with the other method we can check the value of 1/128 = 0.00781 so that agrees quite nicely.