Question

Question #1991a

Answer 1

You'll be left with 6.0% of the initial dose.

Explanation:

So, you know that potassium-42 has a nuclear half-life of 12.4 hours. This means that the initial dose of potassium-42 gets halved every 12.4 hours.

Since the numbers don't allow for a quick calculation, you're going to have to use the nuclear hal-life equation

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

`A(t)` - the amount left after t hours;
`A_0` - the initial dose;;
`t_("1/2")` - the half-life of potassium-42.

The first thing you need to do is convert the time given to you from days to hours

`2.1cancel("days") * "24 hours"/(1cancel("day")) = "50.4 hours"`

This means that you get

`A(t) = A_0 * (1/2)^((50.4cancel("hours"))/(12.4cancel("hours"))`

`A(t) = A_0 * 0.05977`

To get the percentage of the initial dose that remains after 50.4 hours, divide `A(t)` by `A_0` and multiply by 100.

`overbrace(A(t))^(color(blue)("=0.05977" * A_0))/A_0 * 100 = (0.05977 * cancel(A_0))/(cancel(A_0)) * 100 = 5.977%`

Rounded to two sig figs, the answer will be

`"% remaining" = color(green)("6.0%")`