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In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found in
the ice of the Oetztal Alps (hence the name “Oetzi”) in Southern Tyrolia near the Austrian–Italian border, caused
a scientific sensation. When did Oetzi approximately live and die if the ratio of carbon 146C　to carbon　126C in
this mummy is 52.5% of that of a living organism?

1991年9月に有名なアイスマン（オッツィ）、石器時代の新石器時代のミイラ
オーストリアとイタリアの国境に近い南チロルのエッツタールアルプスの氷（「オッツィ」という名前）

このミイラは生物の52.5％ですか？

Physical Information. In the atmosphere and in living organisms, the ratio of radioactive carbon146C (made
radioactive by cosmic rays) to ordinary carbon 126C is constant. When an organism dies, its absorption of 146C
by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive
carbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of 146C , which
is 5715 years

5715年です

Solution. Modeling. Radioactive decay is governed by the ODE y’=ky(see Sec. 1.1, Example 5). By
separation and integration (where t is time and is the initial ratio of 146C to 126C )

解決。モデリング。放射性崩壊はODEによって管理されています（セクション1.1、例5を参照）。沿って

$\frac{dy}{ｙ}= kdt$　、$ln|y|= kt + c$ 、$y= y_0e^kt$ ($y = y_0e^kt$)

Next we use the half-life to determine k. When , half of the original substance is still present. Thus,

$y_0e^{kH}= kdt$, 　$e^{kH} = 0.5$,　$k = \frac{ln0.5}{H} = \frac{0.693}{5715} = -0.0001213$

Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed),

$e^{kt}= e^{-0.0001213t} = 0.525$,　$t= \frac{ln0.525}{-0.0001213}= 5312$

Other methods show that radiocarbon dating values are usually too small. According to recent research, this is
due to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing.

次に、半減期を使用してkを決定します。の場合、元の物質の半分がまだ存在しています。したがって、

$y_0e^{kH}= kdt$, 　$e^{kH} = 0.5$,　$k = \frac{ln0.5}{H} = \frac{0.693}{5715} = -0.0001213$



$e^{kt}= e^{-0.0001213t} = 0.525$,　$t= \frac{ln0.525}{-0.0001213}= 5312$



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