Radioactive 40 K decays to 40 Ar with a halflife of 1.2 x 10^9 years. A sample o
ID: 76526 • Letter: R
Question
Radioactive 40K decays to 40Ar with a halflife of 1.2 x 10^9 years. A sample of moon rock was found tocontain 18% 40K and 82% 40Ar…what isthe age of this rock? t(half-life) = 0.693/k k = 0.693/ 1.2E9years = 5.775E-10/years ln[initial/final] =kt t = ln[initial/final] / k = ln[1/0.180]/ 5.775E-10/ years = 2.969 E 9 years the K starts out initially as 100%(or 1) VERSUS0.693 / 1.2 x 10^9 = 5.775 * 10 ^ -10 yr ^ then i did ln (.82/.81) = 1.516347 1.516347/ 5.775 * 10 ^ -10 yr ^ = 2 26252709938.3 years.
is this right?
t(half-life) = 0.693/k k = 0.693/ 1.2E9years = 5.775E-10/years ln[initial/final] =kt t = ln[initial/final] / k = ln[1/0.180]/ 5.775E-10/ years = 2.969 E 9 years the K starts out initially as 100%(or 1) VERSUS
0.693 / 1.2 x 10^9 = 5.775 * 10 ^ -10 yr ^ then i did ln (.82/.81) = 1.516347 1.516347/ 5.775 * 10 ^ -10 yr ^ = 2 26252709938.3 years.
is this right?
Explanation / Answer
it's ln(1/0.18) because you need the initial and finalconcentrations of ONLY K, not the Ar. the initial concentration ofK is 100%, or 1. Later, we see that the final remainingconcentration is 18% of the remaining, or 0.18. so the equation iswritten ln[1/0.18]
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