Carbon-14 has a half-life of 5730 years. How long will it take 17 grams of carbon-14 to be reduced to 10 grams? Round to the nearest integer.
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17(1/2)^(t/5730) = 10
(1/2)^(t/5730) = 10/17
take log of both sides and use rules of logs
(t/5730) log .5 = log10 - lot17
t/5730 = (1 - log17)/log.5
t = 5730(1 - log17)/log .5 = appr 4387 years
To solve this problem, we can use the concept of exponential decay with the formula:
N(t) = N₀ * (1/2)^(t / T)
Where:
- N(t) is the amount of the substance remaining after time t
- N₀ is the initial amount of the substance
- t is the time that has passed
- T is the half-life of the substance
In this case, we know that the initial amount of carbon-14 is 17 grams, and we want to find the time it takes for it to be reduced to 10 grams. So, let's substitute the given values into the formula:
10 = 17 * (1/2)^(t / 5730)
To solve for t, we can take the logarithm of both sides with a base of 1/2:
log₁/₂(10/17) = t / 5730
Now, we can solve for t:
t = 5730 * log₁/₂(10/17)
Using a calculator to evaluate the right-hand side, we find t ≈ 1972.5. Thus, it will take approximately 1972.5 years for 17 grams of carbon-14 to be reduced to 10 grams.
Finally, rounding to the nearest integer, the answer is approximately 1973 years.