Question #1991a

1 Answer
Jun 19, 2015

Answer:

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%")#