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

Beginner2022-01-03Added 32 answers

The half life of ${}^{\left\{14\right\}}C$ is 5715 years.

The connection between the half life and the decay constant is shown by the equation:$k=\frac{0.693}{{t}_{\frac{1}{2}}}$

$k=\frac{0.693}{5715yr}$

$k=1.21\times {10}^{-4}y{r}^{-1}$

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:

$t=-\frac{1}{k}\mathrm{ln}\frac{{N}_{t}}{{N}_{o}}$

$k=t=-\frac{1}{1.21\times {10}^{-4}y{r}^{-1}}\mathrm{ln}\frac{9.7}{16.3}$

$t-4.28\times {10}^{3}yr$

Paul Mitchell

Beginner2022-01-04Added 40 answers

The half life for ${}^{\left\{40\right\}}K$ decay to ${}^{\left\{40\right\}}Ar$ is $t\frac{1}{2}=1.27\cdot {10}^{9}yrK$

The mass ratio of the rock is ${}^{\left\{40\right\}}Ar$to $}^{\left\{40\right\}}K=\frac{42}{1$

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.

$k=\frac{0.693}{1.27\cdot {10}^{9}yr}$

$=5.46\cdot {10}^{-10}yr\{-1\}$

We can now determine the rock's age.

$\mathrm{ln}(\frac{{N}_{t}}{{N}_{o}})=-kt$

$\mathrm{ln}\left(\frac{1}{52}\right)=-5.46\cdot {10}^{-10}y{r}^{-1}\cdot t-1.649=-5.46\cdot {10}^{-10}y{r}^{-1}\cdot$

$t=\frac{-1.649}{-5.46\cdot {10}^{-10}y{r}^{-1}}$

$=3.02\cdot {10}^{9}yr$

Vasquez

Expert2022-01-07Added 669 answers

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,

$k=\frac{0.693}{t\frac{1}{2}}$

$=\frac{0.693}{(5715yr(\frac{365days}{1yr})(\frac{24hr}{1day})(\frac{60min}{1hr}))}$

$=2.31\times {10}^{-10}mi{n}^{-1}$

the expression "radioactive decay,"

$\mathrm{ln}\frac{N}{{N}_{o}}=-kt$

$\mathrm{ln}\frac{9.7}{17.3}=-(2.31\times {10}^{-10})t$

$t=2.5\times {10}^{9}min$

Result: The age of the shroud is $2.5\times {10}^{9}min$

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