Question #c035b

1 Answer
Nov 30, 2015

#"26 g"#

Explanation:

You don't actually need to use the formula given to you, all you need to know here is that the initial sample of plutonium-239 will be halved with every passing of a half-life.

This means that you can write

#color(blue)(A = A_0 * 1/2^n) " "#, where

#n# - the number of half-lives that pass in a given period of time
#A# - the mass of the sample that remains undecayed
#A_0# - the initial mass of the sample

So, your goal here is to find the value of #n# by using the fact that

#color(blue)(n = "given period of time"/"half-life")#

In your case, the half-life of plutonium-239 is said to be #2.4 * 10^4# years. This means that after #9500# years, #n# will be

#n = (9500 color(red)(cancel(color(black)("years"))))/(2.4 * 10^4 color(red)(cancel(color(black)("years")))) = 19/48#

If the initial sample had a mass of #34.2# grams, then you'll be left with

#A = "34.2 g" * 1/2^(19/48) = "25.994 g"#

Rounded to two sig figs, the answer will be

#A = color(green)("26 g")#