Decay function

[Mrtinkles]

Hello, I am struggling to undertand how to start with this question mainly what goes where and then what to do from there?
For part a the formula is N=N0 × 2^-t/T.
So T = 5568
N0 = 3.00 × 10^-11% ?
How do I know what t is?

The basis for dating archaeological finds is that there has been a constant proportion of atmospheric 14^C. Atmospheric carbon exists as three forms - the stable isotopes 12^C (which constitutes 98.89% of atmospheric carbon) and 13^C (1.11%), and the radioactive form 14^C, which makes up 1.00 × 10^-10 % of the atmospheric carbon. The carbohydrate produced by plants, and consumed by plant- and meat-eaters, has approximately the same composition of the three isotopes of carbon as the atmosphere. But when a plant or animal dies it stops incorporating new carbon and the proportion of 14^C within the soft tissues and skeletons begins to decline.

The half life of 14^C is T = 5568 years, and the radioactive decay of 14C can be modelled using the formula N=N0 × 2^-t/T where t is time.

(a) How old is an archaeological find that has been shown to have 14C equivalent to 3.00 × 10^-11% of the total carbon? Give your answer to 2 sig. figs.


(b) The instruments used to determine the age of an archaeological find are accurate to the percentage of 14C reached after 55000 years. What is this percentage to 2 sig. figs?


(c) Given the decay formula, we should get a straight line by plotting log10N on the y-axis against t on the x-axis. If we were to do this, what would be the value of the intercept of the y-axis and what would the gradient be? Give the units in both cases.

 

[skeeter]

[MATH]N = N_0 \cdot 2^{-t/T}[/MATH]
[MATH]\dfrac{N}{N_0} = 2^{-t/T}[/MATH]
[MATH]\log\left(\dfrac{N}{N_0}\right) = \log\left(2^{-t/T}\right)[/MATH]
[MATH]\log\left(\dfrac{N}{N_0}\right) = (-t/T)\log(2)[/MATH]
[MATH]\dfrac{T}{\log(2)} \cdot \log\left(\dfrac{N}{N_0}\right) = -t[/MATH]
[MATH]\dfrac{T}{\log(2)} \cdot \log\left(\dfrac{N_0}{N}\right) = t[/MATH]
note ...

the original percentage of Carbon 14 is [MATH]N_0 = 1 \times 10^{-10} \text{ %}[/MATH]
the percentage of Carbon 14 after [MATH]t[/MATH] years is [MATH]N = 3 \times 10^{-11} \text{ %}[/MATH]
[MATH]T = 5568[/MATH]

 

[Mrtinkles]

Thanks, what did you do at step 3/4 to get = (-t/T)log(2) ?
and the next step also

 

[skeeter]

Thanks, what did you do at step 3/4 to get = (-t/T)log(2) ?

power property of logs ...

[MATH]\log(a^b) = b\log(a)[/MATH]

and the next step also

algebra ... multiplied both sides by [MATH]T [/MATH] and divided both sides by [MATH]\log(2)[/MATH] to isolate [MATH]-t[/MATH]
After reviewing most of your recent posts on this site, I wonder if you have received any instruction regarding the properties of logarithms and using them to solve exponential equations.

 

[Mrtinkles]

Thanks, I've been watching log videos most of the day.

Are you able to tell me how to work out part c?

 

[skeeter]

Are you able to tell me how to work out part c?

start from the 3rd step in post #2 ...

[MATH]\log\left(\dfrac{N}{N_0}\right) = (-t/T)\log(2)[/MATH]
use the division property of logs on the left side of the equation, then algebraically isolate [MATH]\log(N)[/MATH]

 

[Mrtinkles]

Thanks that's very helpful. I really struggle with maths but it just doesn't come to me like it seems to with others, which is a shame because I actually find it really interesting.

I tried to work it out like this

log( N/N₀) = log(N) - log(N₀)

log(N) - log(N₀) = (-t/T) log(2) ............................ +log(N₀)

log(N) = (-t/T) log(2) + log(N₀)

 

[skeeter]

log( N/N₀) = log(N) - log(N₀)

log(N) - log(N₀) = (-t/T) log(2) ............................ +log(N₀)

log(N) = (-t/T) log(2) + log(N₀)

did you finish up with this last question?

If we were to do this, what would be the value of the intercept of the y-axis and what would the gradient be? Give the units in both cases.