Question

The half life of carbon-14 is 5,730 years. Assume that a typical male has 35 pounds of carbon-14 in his body at death. What is the equation that describes this decay pattern?

Answer 1

it's an exponential decay. Do you know how to write an exponential decay equation? How about convert a half life into a rate constant?

Answer 2

\[C(t)=C _{0} e^{-kt}\]

Answer 3

@dkebler we are learning it without an e, so I was confused when I was trying to look for the equation... only one that I know about it is the y=p(1+r)^x which is the exponential growth equation formula..

Answer 4

now how is k related to half life???? Half life is really just an understandable way to express a rate constant. It' easy for folks to understand that in 5700 years half will be gone.
\[k=\ln2/t_{1/2}\]

Answer 5

it is not necessary to use base e only more convenient.

Answer 6

your (1+r) is the base

Answer 7

r is related to the half life

Answer 8

\[C(t)=C _{0}(1+r)^t\]

Answer 9

now you have to figure out r from the half life

Answer 10

can you do that on your own?

Answer 11

Oh yeah, that's where I got stuck on... to figure out r..

Answer 12

and so, do you use C(t)=C0(1+r)t to solve this question?

Answer 13

ok here is the trick. Half life means time for half to go away

Answer 14

so you divide 2 by times..?

Answer 15

\[1/2(C_{0}=C_{0}(1+r){hl}\]

Answer 16

the Co's cancel you have to take the log of both sides hl is your half life a number you already know then you can solve for r

Answer 17

y=1/2(5730(1+r)^x...?

Answer 18

whoops can't solve for r that way, just a moment

Answer 19

so for example, if I want to figure out how much carbon 14 will be remaining after 1000 years, do I just plug 1000 into x?

Answer 20

just take the hl root of 1/2 and subtract 1

Answer 21

first must find r then you can make a prediction

Answer 22

use the form I gave you X is confusing. t makes more sense. But find r first

Answer 23

But what do I have to do with the 35 pounds.. well it seems most likely related to figure out what r would be.. but I'm not sure about that

Answer 24

a rough estimate

Answer 25

(1/2)^(1/5730)=1+r

Answer 26

r=-0.987 ..?

Answer 27

now you have \[C(t)=35(1+r)^t\]

Answer 28

y=35(1.99)^t, is that what you are trying to say..?

Answer 29

hold on should have been 1-r all along as this was a decay (1+r) must be less than 1.

Answer 30

oh yeah i meant y=35(.01)^t

Answer 31

just a second.

Answer 32

ok I get for the base .9998791

Answer 33

so C(t)=35(.9998791)^t

Answer 34

now you can put in whatever time you want to making predictions.

Answer 35

oh I kind of get it now! but then can you explain how did you find the base .99? because I got different answer from what you got;; I think it's because I used different equation, anyways what kind of formula did you use to find out the base .99? this will be my last question!

Answer 36

the confusion is the 1+r stuff, not needed here the form was just y = p*b^x

Answer 37

so in this case b = (1/2)^(1/5730)=.9998791

Answer 38

so after 1000 years later, there will be .00151of carbon-14..?

Answer 39

no think about it. in 5700 years half is gone, 17.5 remains. In 1000 years it's gonna be like 30 probably

Answer 40

\[C(10000)=35(.9998791)^(10000)\]

Answer 41

that's 1000 not 10000 sorry...getting late for me

Answer 42

oh, I did y=35(.99)^1000. maybe that's why

Answer 43

yes VERY important to keep as many decimal places in the base as possible cause rounding even a bit will throw all answers off significantly because it is an exponential equation. Ok you good?

Answer 44

yup! thank you so much for this hard work^^&