Question

Question #2e0a3

Answer 1

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(t) = A_0 * (1/2)^(t/t_("1/2"))`, where

`A(t)` - the amount left after t years;
`A_0` - the initial quantity of the substance that will undergo decay;
`t_("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(t) = A_0 * 1/100 = A_0/100`

Plug this into the equation to solve for `t/t_("1/2")`

`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

`t/t_("1/2") = log_("1/2")(0.01) = 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.

Read more about the health impact of these isotopes here:

http://www.geigercounter.org/radioactivity/isotopes.htm