the half - life of carbon - 14 is 5600 years. if a piece of charcoal made from the wood of a tree shows only…

the half - life of carbon - 14 is 5600 years. if a piece of charcoal made from the wood of a tree shows only 66% of the carbon - 14 expected in living matter, when did the tree die?\nthe tree died about 3357 years ago.\n(do not round until the final answer. then round to the nearest whole number.)

the half - life of carbon - 14 is 5600 years. if a piece of charcoal made from the wood of a tree shows only 66% of the carbon - 14 expected in living matter, when did the tree die?\nthe tree died about 3357 years ago.\n(do not round until the final answer. then round to the nearest whole number.)

Answer

Explanation:

Step1: Write the formula for carbon - 14 decay

The formula for exponential decay is (A = A_0\left(\frac{1}{2}\right)^{\frac{t}{h}}), where (A) is the current amount, (A_0) is the initial amount, (t) is the time elapsed, and (h) is the half - life. Given (A = 0.66A_0) and (h = 5600).

Step2: Substitute the values into the formula

Substitute (A = 0.66A_0) into (A = A_0\left(\frac{1}{2}\right)^{\frac{t}{h}}), we get (0.66A_0=A_0\left(\frac{1}{2}\right)^{\frac{t}{5600}}). Divide both sides by (A_0), so (0.66=\left(\frac{1}{2}\right)^{\frac{t}{5600}}).

Step3: Take the natural logarithm of both sides

(\ln(0.66)=\ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5600}}\right)). Using the property (\ln(a^b)=b\ln(a)), we have (\ln(0.66)=\frac{t}{5600}\ln\left(\frac{1}{2}\right)).

Step4: Solve for (t)

Since (\ln\left(\frac{1}{2}\right)=-\ln(2)\approx - 0.693) and (\ln(0.66)\approx - 0.416), then (t = 5600\times\frac{\ln(0.66)}{\ln\left(\frac{1}{2}\right)}). [t = 5600\times\frac{- 0.416}{- 0.693}] [t = 5600\times\frac{416}{693}] [t=\frac{5600\times416}{693}] [t=\frac{2329600}{693}\approx3362]

Answer:

The tree died about (3362) years ago.