Asked by Hannah
A bone is found to have 45% of the original amount of carbon 14. If the half life of carbon is 5600 years, how old is the bone.
Would I use the formula N=No e^kt?
Would I use the formula N=No e^kt?
Answers
Answered by
drwls
You could use that formula, if you know how to relate k to half life. In the formula you wrote, there would have to be a negative exponent (assuming k is positive).
Another way that might be easier to remember is to solve:
0.45 = (0.5)^(t/T)
where T is the half life.
Take logs of both sides (to any base)
t/T = log(0.45)/log(0.5) = 1.15
T = 5600/1.15 = 4870 years
Another way that might be easier to remember is to solve:
0.45 = (0.5)^(t/T)
where T is the half life.
Take logs of both sides (to any base)
t/T = log(0.45)/log(0.5) = 1.15
T = 5600/1.15 = 4870 years
Answered by
Reiny
yes, you could
You will need
.5 = 1 e^(5600k)
5600k = ln .5
k = ln .5 / 5600
so .45 = e^((ln.5/5600)t
(ln.5/5600)t = ln.45
t = 5600(ln.45)/ln.5
= 6451.2
or the other formula
N = (.5)^(t/5600)
then .45 = .5^(t/5600)
t/5600 = ln.45/ln.5
t = (5600ln.45/ln.5) = same calculation
You will need
.5 = 1 e^(5600k)
5600k = ln .5
k = ln .5 / 5600
so .45 = e^((ln.5/5600)t
(ln.5/5600)t = ln.45
t = 5600(ln.45)/ln.5
= 6451.2
or the other formula
N = (.5)^(t/5600)
then .45 = .5^(t/5600)
t/5600 = ln.45/ln.5
t = (5600ln.45/ln.5) = same calculation
Answered by
Hannah
Thank You
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