A fossil is found to have a C-14 level of 86% compared to living organisms. How old is the fossil?, (the half life of C14 is 5730 years).
It's a convenient number to pick. You can start with any number you wish as long as you make N = 0.86*No. So it's easy to pick 100% for No which makes N = 86.
The hint they gave us now is that this is a first intigrated order reaction, I do not get it, what the order of the reaction has to do with nuclear chemistry and decay?
The first is the fact that all nuclear decay reactions are first order. That tells you that you can use the ln(No/N) = kt equation and that k = 0.693/t1/2.
Other orders have different equations for t1/2 and for time for x amount to decay. In kinetics it makes little difference if we are talking about decay of a radionuclide or decay of N2 and O2 is a reaction of N2 + O2 ==> 2NO except that radionuclide decay is always first order and other reactions may be second order or third order.
Thanks, I got it right.
To determine the age of the fossil, we need to use the concept of radioactive decay and the equation for exponential decay.
The percentage of C-14 remaining in a fossil compared to living organisms can be used to estimate its age. The half-life of C-14 is 5730 years, which means that after each half-life, the amount of C-14 will reduce by half.
Let's break down the question:
Initially, the fossil is said to have a C-14 level of 86% compared to living organisms. This means that the fossil still has 86% of the initial amount of C-14 present.
We need to find the number of half-lives that have occurred to reach the current C-14 level of 86%. To do this, we can use the equation:
Final amount = Initial amount × (1/2)^(number of half-lives)
Rearranging the equation, we get:
(number of half-lives) = log(final amount / initial amount) / log(1/2)
Plugging in the given values:
(final amount) = 86% = 0.86
(initial amount) = 100% = 1
So, the number of half-lives can be calculated as:
(number of half-lives) = log(0.86) / log(1/2)
Using a scientific calculator, evaluating this expression, we find that the number of half-lives is approximately 0.251.
To find the age of the fossil, we multiply the number of half-lives by the half-life of C-14:
Age of the fossil = (number of half-lives) × (half-life of C-14)
Age of the fossil = 0.251 × 5730 years
Evaluating this expression, we find that the age of the fossil is approximately 1439 years.
k = 0.693/t1/2
ln(No/N) = kt
I would use 100 for No
and N = 86
k from above
solve for t in years.