Archeologists can determine the age of artifacts made of wood or bone by measuring the amount of the radioactive slope 14C present in the object. The amount of isolope decreases in a first-ordrr process. If 12.8% of the original amount of 14C is present in a wooden tool at the time of analysis, what is the age of the tool? The half-life of 14C is 5730 yrs.
k = 0.693/t1/2
Then substitute k in the following:
ln(No/N) = kt
No = 100 (I assume this arbitrarily and calculate N based on what I chose for N.)
N = 12.8
k from above
t = solve for t in years.
Note the correct spelling of isotope.
To determine the age of the wooden tool, we can use the concept of radioactive decay. The half-life of an isotope is the time it takes for half of the original amount of that isotope to decay.
In this case, the half-life of 14C is given to be 5730 years. This means that every 5730 years, the amount of 14C in the wooden tool decreases by half. We can use this information to find the age of the tool.
Let's assume that the original amount of 14C in the wooden tool was 100%. According to the problem, 12.8% of the original 14C is still present. This means that 87.2% has decayed.
To find the number of half-lives that have passed, we can use the decay formula:
Nt = N0 * (1/2)^(t/t_half)
Where:
- Nt is the amount of remaining 14C (12.8% in this case)
- N0 is the original amount of 14C (100%)
- t is the time that has passed in years
- t_half is the half-life of the isotope (5730 years)
By rearranging the formula, we can solve for t:
t = t_half * (log(Nt/N0) / log(1/2))
Plugging in the given values:
t = 5730 * (log(0.128/1) / log(1/2))
Using a scientific calculator, we can find that log(0.128/1) ≈ -0.894.
t = 5730 * (-0.894 / log(1/2))
log(1/2) = -0.301 (approximately).
t ≈ 5730 * (-0.894 / -0.301)
t ≈ 5730 * 2.973
t ≈ 17052.6
Therefore, the age of the wooden tool is approximately 17053 years.