Help with a C-14 Half Life question!!?

Question

Hi. I am completely stuck on a question i am trying to solve. It seems it should be simple. Im sure it probably is but i just cant get my head around it. Here it is!

An object taken from a cave has a carbon-14 fraction which is 0.89 of the amount in a living organism. How old is the object? (half-life for C-14 is 5730 yr)

Im not looking for the answer as i already know what it is. But can anyone show me how to go about working this out. If you could explain each mathematical step that would be amazing.

Thanks in advanced.

Nick

Answer 1

Best Answer - Chosen by Voters

sorry dude ......it is a chemistry question ........if u give a formula, i can solve it.

Answer 2

Well if 50% or 0.50 was left, we know we have had 1 half-life meaning an age of 5730 years.

We have 0.89, which is 0.11 less than 1 and we need to calculate what %age this is of the half-life.

0.11 / 0.50 = 0.22 of a half-life.

5730 * 0.22 = 1260.6 years or 1260 years, 7 months and 6 days.

Answer 3

if N(t) is the amount at time t
N(0) is the initial amount
N(t) = N(0) (1/2)^( t /Half-Life)

0.89 = 1 (1/2)^(t/(5730)
log( .89 )) = (t / ( 5730) log(1/2)
t = 5730 log( .89 ) / log(1/2) = 963.343 years

Answer 4

I will use th for half life.

Basic expression for the meaning of half life
0.89 = (1/2)^(t/th)

take natural log of both sides
ln 0.89 = (t/th) ln (1/2)

Solve for t
t = th * ln(0.89) / ln(1/2) = 5730 * (-0.1165) / (-0.6931) = 963 yrs