Question

HELP ME PLEASE!!
An artifact originally had 16 grams of Carbon-14 present. Determine how many grams of Carbon-14 will be present in 11,430 years.

Answer 1

i think you are suppose to up A=Pe^rt

Answer 2

do you know how to solve it @ganeshie8 ??

Answer 3

yes

Answer 4

they gave rate of decay ?
or
they expect us to know half life of C-14 and proceed ?

Answer 5

well they didnt give rate or say its a half life so i didnt know what to do. in the question before this it asked me to find the half life of it but thats it

Answer 6

ok. then we can use half-life of C-14 here

can u google and tell me halflife of C-14

Answer 7

wait a sec

Answer 8

yep.. we need to use halflife here,
from that we can compute rate-of-decay

Answer 9

it said it was 5,730 years

Answer 10

\(\huge \frac{1}{2} = e^{r*5730}\)

Answer 11

right ?

Answer 12

yes

Answer 13

great !
take log both sides, and compute "r"

Answer 14

once you have "r",

you can use the formula
\(\huge P = Ae^{rt}\)

Answer 15

can you try ?

Answer 16

ias a final answer i got 1,591,056

Answer 17

whats that ?

16 grams decayed to 1.5 million grams ?????

hows that possible

Answer 18

whats the value of "r" ?

Answer 19

its not decayed its growth

Answer 20

lol yes hows that possible

Answer 21

the question is after 11,430 years how many grams of c-14 will there be

Answer 22

u mean,
the question is asking to find 11,430 years back how much quantity of C-14 was present ?

Answer 23

from the halflife equation, you should get a -ve value for "r"

Answer 24

if you go back in time, you will put -ve value for time

Answer 25

then it can become growth. but i guess question is not asking for that

Answer 26

the question is simply asking how many grams of Carbon - 14 will be present in 11,430 years if the artifact originally had 16 grams

Answer 27

@agentx5 if you're familiar with this... pls help

Answer 28

Ok, I see what happened here:

First, @ganeshie8 , did you get \(\huge-\frac{ln|2|}{5730}\) for r?

Answer 29

(I'm assuming you all have the correct half-life written down btw)

Answer 30

yes 5730 years

Answer 31

@agentx5 i did get -ve for r. and i know the quantity should decay

Answer 32

maybe Lulu if you can show ur work how you got 1.59 million it helps

Answer 33

\(\huge A = 4 \sqrt[191]{2} \approx 4.0145 \ \text{g}\) You must be making a calculation error..

Answer 34

This makes logical sense too, because it's approximately twice the half-life time so you would expect another half would be gone.

Answer 35

yeah makes sense to me too :)

Answer 36

You will have approximately half of or half the Carbon 14 you had 11430 years earlier (16g originally)

Answer 37

Very rough approximation to know if you're in the right ballpark so to speak:
\[16 g \cdot ( \frac{1}{2} * \frac{1}{2}) = 16g \cdot (\frac{1}{4}) = 4\]

Answer 38

\[\frac{5730 \ yrs}{11430 \ yrs} \approx \frac{1}{2}\]

Answer 39

If this is a multiple choice exam I would get a little practice in understand the concept and the logic I'm using here to test what the answer should be around before going exact. Unless they professor has two very close numbers on the exam you might even be able to make an educated guess and save yourself a lot of time. You should also know how to work the formulas.

The answer with the 191st root above is in fact the perfectly exact answer, and if it was a written exam that's what I would write, with units of grams.

Answer 40

Make sense @Lulu212 ?

Answer 41

ok.

Answer 42

but this is not an exam its hw

Answer 43

Example: if my choices were...
A) 10g
B) 3g
C) 4g
D) 6g

I could just use reasoning to figure out the answer should be damn close to 4 and on the little bit more side.

Answer 44

Homework is practice for exams ;-)

Answer 45

And if I was teaching Chemistry 1, you better expect I would have half-life on the exam. It's just one of those fundamental concepts you should know in chemistry.

Answer 46

but how is it possible to go from 16 to 4 in 11430 if it meant to increase??

Answer 47

Graph of the amounts present with respect to time. C\(^{14}_6\) decays into Carbon-12. At first quickly, then starting to slow down.