Question

a mummy is found to be 3940 years old by ^14C dating. Using half life of 5730 years for ^14C demonstrate that 62.1% of ^14C at the death of the person remains the mummy

Answer 1

For 1st order reactions you can use these formulas:

find the decay constant

\[t _{1/2}=\frac{ \ln2 }{ k }\]

t1/2 = half-life

then use the general exponential decay/growth equation:

\[A _{t}=A _{0}*e ^{-kt}\]

Ao=initial amount
At= amount after time elapsed
t=time elapsed
k=decay constant

start off by assuming you have 100% 14C, i.e. Ao=100

Answer 2

ok then what

Answer 3

.125=e^-k(5730)

Answer 4

you set it up wrong

Answer 5

find k first

Answer 6

how?

Answer 7

with the first equation i wrote

Answer 8

what would At be though

Answer 9

okay rewind a little bit, its asking you to prove that 62.1% remains of the mummy. now you can do this several ways, find what At is assuming all other variables OR use 62.1% for At and find the initial amount Ao.

Answer 10

im sorry I still don't understand if Ao equals 100 then what would At and t be

Answer 11

well if you're using Ao as 100% then you're solving for At

if you're using At as 62.1% then you're solving for Ao

Answer 12

t is time elapsed so 3940 years

Answer 13

what is the k value we are using

Answer 14

you found it in the first equation using the half life

Answer 15

ok then I got 1.21x10^-4 for k

Answer 16

use it in the other equation

Answer 17

i did and i got .621 which is 62.1% right?

Answer 18

yep, in decimal, 0.621 is 62.1%

Answer 19

ok thanks so much
do you have time for one more?

Answer 20

no problem, yeah, what is it?

Answer 21

nevermind i figured it out thanks so much i was so screwed because i have test tomorrow

Answer 22

okay, good luck on your test !