Question

The burial cloth of an Egyptian mummy is estimated to contain 560 g of the radioactive materialcarbon-14, which has a

Exponential models
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asked 2021-02-12

The burial cloth of an Egyptian mummy is estimated to contain 560 g of the radioactive materialcarbon-14, which has a half life of 5730 years.
a. Complete the table below. Make sure you justify your answer by showing all the steps.
\(\begin{array}{|l|l|l|}\hline t(\text{in years})&m(\text{amoun of radioactive material})\\\hline0&\\\hline5730\\\hline11460\\\hline17190\\\hline\end{array}\)
b. Find an exponential function that models the amount of carbon-14 in the cloth, y, after t years. Make sure you justify your answer by showing all the steps.
c. If the burial cloth is estimated to contain 49.5% of the original amount of carbon-14, how long ago was the mummy buried. Give exact answer. Make sure you justify your answer by showing all the steps.

Answers (1)

2021-02-13

Given that
Initially the radioactive carbon-14 present in the burial cloth of the egyptian mummy is 560g.
The half life of carbon-14 is 5730 years.
The total amount of carbon-14 decayed after t years is given by
\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{1}}}{{{h}}}}}}\) (1)
Where, \(\displaystyle{A}_{{0}}\) initial amount of carbon -14
t-time in years,
h-Half life of carbon -14
a) To complete table using the following calculation:
Now,
1) t=0 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g
\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{1}}}{{{h}}}}}}\)
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{0}}}{{{5730}}}}}}\)
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{0}}={560}{\left({1}\right)}\)
\(\displaystyle{A}={560}\)g
2) \(t=5730\) years, \(h=5730\) years,\(\displaystyle{A}_{{0}}={560}\)g
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{5730}}}{{{5730}}}}}}\)
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{1}}={560}{\left({0.5}\right)}\)
\(\displaystyle{A}={280}\)g
3) \(t=11460\) years, \(h=5730\) years,\(\displaystyle{A}_{{0}}={560}\)g
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{11460}}}{{{5730}}}}}}\)
\(\displaystyle{560}{\left({0.5}\right)}^{{2}}={560}{\left({0.25}\right)}\)
\(\displaystyle{A}={140}\)g
3) \(t=17190\) years, \(h=5730\) years,\(\displaystyle{A}_{{0}}={560}\)g
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{17190}}}{{{5730}}}}}}\)
\(\displaystyle{A}={560}{\left({0.5}\right)}^{{3}}={560}{\left({0.125}\right)}\)
\(\displaystyle{A}={70}\)g
The table becomes
\(\begin{array}{|l|l|l|}\hline t(\text{in years})&m(\text{amoun of radioactive material})\\\hline0&560\\\hline5730&280\\\hline11460&140\\\hline17190&70\\\hline\end{array}\)
b)To find the exponential function that models the amount of carbon-14 in the cloth, y, after t years:
\(\displaystyle{A}={A}_{{0}}{e}^{{-{k}{t}}}\)
Here, At \(t=0, y=560\) g,
k- rate of decay,
t- time in years.
The exponential function is
\(\displaystyle{y}={560}{e}^{{-{k}{t}}}\)
c) To find t, for \(k=49.5\%\) :
49.5% of original amount of carbon-14 is
\(\displaystyle{49.5}\%\times{560}={\frac{{{49.5}}}{{{100}}}}\times{560}\)
\(\displaystyle={49.5}\times{5.6}\)
\(\displaystyle{49.5}\%\times{560}={277.2}\)g
Now, \(A_{0} =560\), \(A= 277.2\) g, \(h=5730\) years, from (1)
\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{t}}}{{{h}}}}}}\)
\(\displaystyle{277.2}={560}{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}\)
\(\displaystyle{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}={\frac{{{277.2}}}{{{560}}}}\)
\(\displaystyle{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}={0.495}\)
\(\displaystyle{\frac{{{t}}}{{{5730}}}}{\ln{{0.5}}}={\ln{{0.495}}}\)
\(\displaystyle{t}={5730}{\left({\frac{{{\ln{{0.495}}}}}{{{\ln{{0.5}}}}}}\right)}\)
\(\displaystyle{t}={5730}{\left({\frac{{-{0.7032}}}{{-{0.693}}}}\right)}\)
\(\displaystyle{t}={5730}{\left({1.015}\right)}\)
\(\displaystyle{t}={5814.33}\) years
Thus, the mummy was burried before 5814.33 years.

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