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# Calculating when the man died?

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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?

Feb 20, 2020

#1
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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.   Feb 20, 2020
#2
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0.527 = 1/2^(t/5730)

t/5730 =Log(0.527) / log(1/2)

t =5730 x 0.924125 =5295.24 years ago when he died.

5295.24 - 1991 =3304.24 = ~3,300 BC - when he died.

Feb 20, 2020