# Question 1991a

Jun 19, 2015

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

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

$A \left(t\right)$ - the amount left after t hours;
${A}_{0}$ - the initial dose;;
${t}_{\text{1/2}}$ - 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"))

$A \left(t\right) = {A}_{0} \cdot 0.05977$

To get the percentage of the initial dose that remains after 50.4 hours, divide $A \left(t\right)$ 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%")#