SOLUTION: Hello, I am having trouble with this question - could you please provide me with the steps, and solution? If the half-life of Carbon 14 is 5730 years, how muc will remain of

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hello, I am having trouble with this question - could you please provide me with the steps, and solution? If the half-life of Carbon 14 is 5730 years, how muc will remain of       Log On


   

Question 289033: Hello,
I am having trouble with this question - could you please provide me with the steps, and solution?
If the half-life of Carbon 14 is 5730 years, how muc will remain of an initial 3 gms after 1000 years?
Thanks!
Found 4 solutions by mananth, ikleyn, josgarithmetic, greenestamps:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
Half life formula is given by
A=y(1/2)^t/h
A= 3(1/2)^1000/5730
A= 2.6581

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.

If to keep four decimals after the decimal point, then
the correct answer is 2.6582 grams instead of 2.6581 in the post by @mananth.



Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
If model is of the form p=ae%5E%28-kt%29
then for the given half-life
1%2F2=1%2Ae%5E%28-k%2A5730%29
ln%281%2F2%29=-5730%2Ak
k=%28ln%281%2F2%29%29%2F%28-5730%29
k=0.000121

To use the formula and the other given information
p=3%2Ae%5E%28-0.000121%2A1000%29
p=3%2Ae%5E%28-0.121%29
p=%280.886%29%283%29
highlight%28p=2.66%29

Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Using the definition of half life, the amountA%28n%29 remaining after n half-lives is A%28n%29=A%280%29%28%281%2F2%29%5En%29, where A%280%29 is the initial amount.

In this problem, the number of half-lives is 1000%2F5730.

With an initial amount of 3 grams, the amount remaining after that many half-lives is

A%283%29=%283%29%28%281%2F2%29%5E%281000%2F5730%29%29

= 2.658186689....

Remember that radioactive decay is a statistical process; that mathematical answer is only approximately correct