Chris Cruz

2022-01-02

The cloth shroud from around a mummy is found to have a ${}^{\left\{14\right\}}C$ activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for ${}^{\left\{14\right\}}C$ decay, 5715 yr, calculate the age of the shroud.

veiga34

The half life of  is 5715 years.
The connection between the half life and the decay constant is shown by the equation:$k=\frac{0.693}{{t}_{\frac{1}{2}}}$
$k=\frac{0.693}{5715yr}$
$k=1.21×{10}^{-4}y{r}^{-1}$
The information provided lets us know that ${N}_{0}=16.3$ and ${N}_{t}=9.7$
Use the equation $\mathrm{ln}\frac{{N}_{t}}{{N}_{o}}=-kt$
Rearrange to determine the shroud's age:
$t=-\frac{1}{k}\mathrm{ln}\frac{{N}_{t}}{{N}_{o}}$
$k=t=-\frac{1}{1.21×{10}^{-4}y{r}^{-1}}\mathrm{ln}\frac{9.7}{16.3}$
$t-4.28×{10}^{3}yr$

Paul Mitchell

The half life for  decay to  is $t\frac{1}{2}=1.27\cdot {10}^{9}yrK$
The mass ratio of the rock is to
Hence, ${N}_{o}=4.2+1=5.2$
And, ${N}_{t}=1$
Let us calculate the age of the rock.
Let's first determine the decay constant.
$k=\frac{0.693}{1.27\cdot {10}^{9}yr}$
$=5.46\cdot {10}^{-10}yr\left\{-1\right\}$
We can now determine the rock's age.
$\mathrm{ln}\left(\frac{{N}_{t}}{{N}_{o}}\right)=-kt$
$\mathrm{ln}\left(\frac{1}{52}\right)=-5.46\cdot {10}^{-10}y{r}^{-1}\cdot t-1.649=-5.46\cdot {10}^{-10}y{r}^{-1}\cdot$
$t=\frac{-1.649}{-5.46\cdot {10}^{-10}y{r}^{-1}}$
$=3.02\cdot {10}^{9}yr$

Vasquez

We know:
The half time for decay, $t\frac{1}{2}=5715yr$
The disintegration per minute per gram of carbon, N=9.7
The disintegration per minute per gram, ${N}_{o}=16.3$
The rate constant's expression,
$k=\frac{0.693}{t\frac{1}{2}}$
$=\frac{0.693}{\left(5715yr\left(\frac{365days}{1yr}\right)\left(\frac{24hr}{1day}\right)\left(\frac{60min}{1hr}\right)\right)}$
$=2.31×{10}^{-10}mi{n}^{-1}$
$\mathrm{ln}\frac{N}{{N}_{o}}=-kt$
$\mathrm{ln}\frac{9.7}{17.3}=-\left(2.31×{10}^{-10}\right)t$
$t=2.5×{10}^{9}min$
Result: The age of the shroud is $2.5×{10}^{9}min$

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