Here is the problem:
Carbon-14 has a half-life of approximately 5730 years. It is often used to date organic materials that contain carbon. When the concentration of carbon-14 in an organic sample is known, it can be used to calculate the date of the material. "Ötzi,"a mummified man, was found in 1991 in the Ötzal Alps, on the border between Austria and Italy. He is Europe's oldest known natural human mummy. The concentration level of carbon-14 found in his garments was approximately 52.7%. To the nearest hundred years, when did the man die?
I'm not really sure how to approach this question. I'm thinking of using the formula A = A_0e^(kt) where it represents the amount of any item under continuous growth, where A_0 is the initial amount, t is the time period, and k is the growth factor. But it doesn't make much sense?
The answer is 3300 BC, how should I get this number?
+128473 We can find k as follows :
Call the intial amt, C
And the amount left after 5730 yrs must be (1/2)C
So....we have this equation
(1/2)C = C * e ^(k * 5730)
(1/2) = e^(k * 5730) take the Ln of both sides
Ln (1/2) = Ln e ^(5730k) and we can write
Ln (1/2) = (5730k) Ln e [ Ln e = 1....so we can ignore this ]
So we have
Ln (1/2) = 5730 k divide both sides by 5730
Ln (1/2) / 5730 = k = Ln (.5) / 5730
So....to solve the problem ....we have that
.527 = e^(Ln (.5)/5730 * t) take the Ln of both sides
Ln (.527) = Ln e^( Ln(.5)/5730 * t)
Ln (.527) = Ln (.5)/5730 *t
t = Ln (.527) / [ Ln(.5) / 5730]
t = 5730 * Ln (.527) / Ln (.5) = 5295. 24 years old
So.....the man died @ absolute value [1991 - 5295.24] B.C. ≈ 3304.24 B. C. ≈ 3300 B. C.
