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?
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 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
Thank You
Yes, you can use the formula N = Nₒ * e^(-kt) to solve this problem. In this formula:
- N is the current amount of carbon-14 remaining in the bone,
- Nₒ is the original amount of carbon-14 in the bone,
- k is the decay constant (ln(2) / t₁/₂),
- t is the time (in this case, the age of the bone), and
- t₁/₂ is the half-life of carbon-14.
To find the age of the bone, we need to rearrange the formula and solve for t:
t = (ln(N/Nₒ)) / -k
Given that the bone has 45% of the original amount of carbon-14, we can say:
N = 0.45 * Nₒ
Substituting this value into the formula, we have:
t = (ln(0.45 * Nₒ / Nₒ)) / -k
Simplifying further:
t = (ln(0.45)) / -k
Finally, we need to substitute the value of k, which is given by k = ln(2) / t₁/₂, where t₁/₂ is the half-life of carbon-14 (5600 years in this case):
t = (ln(0.45)) / -(ln(2) / 5600)
Evaluating this expression will give you the age (t) of the bone.
Well, we could use that formula, but let me explain it in a more humorous way.
Imagine the bone as a fossilized comedian who loves carbon dating jokes. So, initially, the bone had a full set of jokes, represented by 100% of the original amount of carbon-14.
But as time passed, the bone's jokes became less funny. So now, it only has 45% of its original comedic ability, or in scientific terms, carbon-14.
Now, the half-life of carbon-14 is 5600 years, which means it takes 5600 years for the bone's jokes (carbon-14) to lose half of their humor.
Using a little math magic, we can figure out how many half-lives the bone has gone through to reach 45%. If each half-life is 5600 years, and the bone has 45% left, then it must have gone through two half-lives, losing half of its humor each time.
So, multiplying 5600 years by 2, we find that the bone is approximately 11,200 years old. That's quite an ancient comedian!