Skeletal remains had lost 84% of the C-14 they originally contained. Determine the approximate age of the bones. (Assume the half life of carbon-14 is 5730 years. Round your answer to the nearest whole number.)
Oh, bones that have lost their charm! Let's see if we can find their approximate age by doing a little math and comedy.
Since the half-life of carbon-14 is 5730 years, we can assume that after one half-life, the bones would have lost 50% of their C-14 content. But in this case, they have lost a whopping 84%! That means they have gone through more than one half-life.
To find out how many half-lives it took for the bones to lose 84% of their C-14, we can use the formula:
(Percent remaining / 100) = (1/2)^(number of half-lives)
Substituting the values we have:
(84 / 100) = (1/2)^(number of half-lives)
To solve this comedic equation, we take the logarithm of both sides (log base 1/2), which gives us:
log (84/100) = number of half-lives
Now, let me consult my trusty calculator... *beep boop beep* ...and I find that the number of half-lives is approximately 3.076.
Since each half-life is 5730 years, we can multiply that by 3.076 to get the approximate age of the bones:
5730 years/half-life * 3.076 half-lives = 17,370 years.
Therefore, with a twist of mathematical humor, it appears that the bones are approximately 17,370 years old!
To determine the approximate age of the bones, we can use the formula for exponential decay:
Nt = N0 * (1/2)^(t / T)
Where:
- Nt is the remaining amount of C-14
- N0 is the original amount of C-14
- t is the time that has passed
- T is the half-life of C-14 (5730 years in this case)
In this problem, Nt/N0 is given as 0.16 (100% - 84%) since 84% of C-14 has been lost. Therefore, we can rewrite the formula as:
0.16 = (1/2)^(t/5730)
Now, let's solve for t:
Take the logarithm of both sides to eliminate the exponent:
log(0.16) = log((1/2)^(t/5730))
Using the property of logarithms (log(a^b) = b * log(a)):
log(0.16) = (t/5730) * log(1/2)
Solving for t:
t/5730 = log(0.16) / log(1/2)
t = 5730 * (log(0.16) / log(1/2))
Using a calculator, we can find that log(0.16) / log(1/2) ≈ 5.660964
t ≈ 5730 * 5.660964 ≈ 32485.49
Rounding this value to the nearest whole number, we get:
t ≈ 32485
Therefore, the approximate age of the bones is 32485 years.
To determine the approximate age of the bones, we can use the concept of radioactive decay and the half-life of carbon-14.
The half-life of carbon-14 is the time it takes for half of the original amount of carbon-14 in a sample to decay. In this case, the half-life of carbon-14 is 5730 years.
Since the skeletal remains have lost 84% of the carbon-14 they originally contained, it means that only 16% (100% - 84%) of the original carbon-14 is left.
Now, let's set up an equation to solve for the age of the bones. We know that the remaining carbon-14 is equal to 16% of the original carbon-14.
Remaining carbon-14 = Original carbon-14 * (16/100)
Since the remaining carbon-14 is related to the original amount after a certain number of half-lives, we can express this as:
Remaining carbon-14 = Original carbon-14 * (1/2)^(number of half-lives)
Setting these two equations equal to each other, we have:
Original carbon-14 * (16/100) = Original carbon-14 * (1/2)^(number of half-lives)
We can simplify this equation by canceling out the original carbon-14 on both sides:
(16/100) = (1/2)^(number of half-lives)
To solve for the number of half-lives, we can take the logarithm (base 1/2) of both sides:
log(16/100) = log((1/2)^(number of half-lives))
Using the property of logarithms, we can bring the exponent down:
log(16/100) = number of half-lives * log(1/2)
Now, we can solve for the number of half-lives:
number of half-lives = log(16/100) / log(1/2)
Using a calculator, we find that the number of half-lives is approximately 3.897.
Since each half-life is 5730 years, we can multiply the number of half-lives by the half-life duration to find the age of the bones:
Age of the bones = 3.897 * 5730
Rounding this value to the nearest whole number, we get the approximate age of the bones:
Age of the bones ≈ 22296 years
(1/2)^t/5730 = .16
t/5730 = log(.16)/log(.5)
t = 5730*log(.16)/log(.5)
t = 15149
makes sense. 12.5% left would be three half-lives, or 17190 years