Mummy Carbon Dating

The half-life formula is $A = A_0(\frac{1}{2})^{\frac{t}{h}}$ , where $A$ is the final amount, $A_0$ is the starting amount, $t$ is the time, and $h$ is the half-life.

After $2000$ years, the scroll will only have $(\frac{1}{2})^{\frac{2000}{6000}}$ of its original $\ce{^{14}C}$ remaining.

Since the mummy has $\frac{2}{3}$ of that, we have $\frac{2}{3}(\frac{1}{2})^{\frac{2000}{6000}} = (\frac{1}{2})^{\frac{x}{6000}}$ .

Solving for $x$ gives $x = \frac{6000 \ln (\frac{2}{3}(\frac{1}{2})^{\frac{1}{3}})}{\ln(\frac{1}{2})} \approx \boxed{5509.8}$ .

The half-life formula is given by $A = A_0 (\frac{1}{2}) ^\frac{t}{h}$ , where $A$ is the final amount, $A_0$ is the initial amount, $t$ is the time that has passed, and $h$ is the half-life of the item being measured.

Using subscripts to denote different samples,

$A_1 = A_0 (\frac{1}{2}) ^\frac{t_1}{h}$ and $A_2 = A_0 (\frac{1}{2}) ^\frac{t_2}{h}$

Let $k$ be the proportion of $\ce{^12C}$ between the samples. Hence, $A_1 = kA_2$

Equating and simplifying, we end with the relationship $t_1=t_2-\frac{h\ln k}{\ln 2}$ . We can now compare any 2 samples.

Plugging in the values $t_2=2000$ , $h=6000$ and $k= \frac{2}{3}$ ,

$t_1=2000-\frac{6000 \ln \frac{2}{3}}{\ln 2}\approx \boxed{5509.8}$

All the answers are good. I tend to use exponential functions to work with decay, which is basically just a notation preference. You can model exponential decay with the formula $N(t) = N_0 e^{-t/\tau}$ where $N(t)$ is the number of $^{14} C$ atoms at time $t$ , $N_0$ is the number at $t=0$ , and $\tau$ is the exponential time constant. We are given that $\frac{N_0}{2} = N_0 e^{-6000/\tau}\ \Rightarrow ln(\frac{1}{2}) = -\frac{6000}{\tau}\ \Rightarrow \tau = -\frac{6000}{ln(\frac{1}{2})}$ . We are then given that $\frac{2}{3} e^{-2000/\tau} = e^{-x/\tau}\ \Rightarrow -\frac{x}{\tau} = ln(\frac{2}{3} e^{-2000/\tau}) = -\frac{2000}{\tau} + ln(\frac{2}{3})\ \Rightarrow x = 2000 - ln(\frac{2}{3})\tau = 2000 + 6000\frac{ln(\frac{2}{3})}{ln(\frac{1}{2})} \approx 5510$ .

log $\frac{1}{2}$ ( $\frac{2}{3}$ ) * 6000 + 2000 = 5510 years

I've been out of school for a while so here's how I did it:

1) Let A be the exponential decay constant, such that after one half-life exactly one-half of the 14C remains.

2) Let B be the number of years necessary for exactly two-thirds of the original amount of 14C to remain.

Therefore from the problem we can state:

$A^{6000} = \frac{1}{2}$ , and

$A^{B}= \frac{2}{3}$

We can take the log of both sides to get

$6000log{A}=log{\frac{1}{2}}$ , and

$Blog{A}=log{\frac{2}{3}}$

We solve the system for B to get:

$B=\frac{6000\cdot log{\frac{2}{3}}}{log{\frac{1}{2}}}\approx3510$ years.

This is how much older the mummy is than the scroll. The scroll is already known to be 2000 years old and so this makes the mummy 5510 years old.

Just solve 2^(-x/6000)=2/3*2^(-y/6000), for y=2000

Let us define $S_{\ce{C}}$ and $M_{\ce{C}}$ as the scroll's and mummy's respective $\ce{^{14}C}:\ce{^{12}C}$ ratios.

The function for the radioactive decay of $\ce{^{14}C}$ , which is at the same time the function for the ratio of $\ce{^{14}C}:\ce{^{12}C}$ , is $N(t) = N_0 * x^t$ , where $t$ is the time given in years. The starting value $N_0$ is unimportant as the problem concerns ratios, and $x$ is unknown.

We know that $x^{6000}=\frac{1}{2}$ and can from there reason that:

$x = \sqrt[6000]{\frac{1}{2}}$

$x =\frac {1} {\sqrt[6000]{2}}$

$x = {2}^ {-\frac{1}{6000}}$

This makes the formula for radioactive decay $N(t) = {2}^ {-\frac{t}{6000}}$

$S_{\ce{C}} = N({2000}) = {2}^ {-\frac{2000}{6000}} = \frac {1}{\sqrt[3]{2}}$

For the age $a$ of the mummy, we can take the equation:

$M_{\ce{C}}= {2}^{-\frac{a}{6000}} = \frac23 * \frac{1}{\sqrt[3]{2}}$

$a * \ln({2}^{-\frac{1}{6000}}) = \ln(\frac {2}{{3}\sqrt[3]{2}})$

$a = \frac{\ln(\frac {2}{{3}\sqrt[3]{2}})}{\ln({2}^{-\frac{1}{6000}})} = 5509.775\ldots \approx \boxed {5510}$

log(1/2)/log(2/3)=3510

3510+2000=5510 years.