# Question 2e0a3

Mar 9, 2015

It will take 54 days for iodine-131 and 193 years for strontium-90 to reach 1% of their initial values.

Once again, your tool for solving this problem is the nuclear half-life equation

$A \left(t\right) = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\text{1/2}\right)}$, where

$A \left(t\right)$ - the amount left after t years;
${A}_{0}$ - the initial quantity of the substance that will undergo decay;
${t}_{\text{1/2}}$ - the half-life of the decaying quantity.

In both cases, you can write the amount left after the passing of $t$ years by using the initial amount, ${A}_{0}$. Since the activity of these isotopes must fall to 1% of its initial value, only the 100th part of the initial value will remain

$A \left(t\right) = {A}_{0} \cdot \frac{1}{100} = {A}_{0} / 100$

Plug this into the equation to solve for $\frac{t}{t} _ \left(\text{1/2}\right)$

A_0/100 = A_0 * (1/2)^(t/t_("1/2")) => 1/100 = (1/2)^(t/t_("1/2")

Taking the base 1/2 log from both sides will yield

$\frac{t}{t} _ \left(\text{1/2") = log_("1/2}\right) \left(0.01\right) = 6.644$

In the case of iodine-131, the value for its half-life is 8.1 days, which means $t$ is equal to

t_("iodine") = t_("1/2") * 6.644 = "8.1 days" * 6.644 = "53.8 days"

Rounded to two sig figs, the number of sig figs in 8.1, the answer will be

t_("iodine") = "54 days"

In the case of strontium-90, the value for its half-life is 29.1 years, which means $t$ will be equal to

t_("strontium") = "29.1 years" * 6.644 = "193.3 years"

Rounded to three sig figs, you'll have

t_("strontium") = "193 years"#

As you can see, the strontium-90 isotope poses far more long term dangers; in fact, traces of strontium-90 can be found in the teeth of all the people born after 1963 because of the extensive nuclear testing done before that date.