Question

1. the oldest bone in the human body is the femur. at an archeological site, a femur bone was found. the original amount of ¹⁴c was estimated to be 64 g. the present amount of ¹⁴c was found to be 4 g. how old is the bone? carbon 14 (¹⁴c) has a half - life of 5760 years. a = a₀(1/2)^(t/h) g: a = 4g a₀ = 64g h = 5760 r: t

Explanation:

Step1: Identify the decay formula

The formula for radioactive decay is $A = A_0(\frac{1}{2})^{\frac{t}{h}}$, where $A$ is the present - amount, $A_0$ is the initial amount, $t$ is the time elapsed, and $h$ is the half - life. Given $A = 4$ g, $A_0=64$ g, and $h = 5760$ years.

Step2: Substitute the values into the formula

Substitute the values into the formula: $4=64(\frac{1}{2})^{\frac{t}{5760}}$. First, divide both sides of the equation by 64: $\frac{4}{64}=(\frac{1}{2})^{\frac{t}{5760}}$, which simplifies to $\frac{1}{16}=(\frac{1}{2})^{\frac{t}{5760}}$.

Step3: Rewrite the left - hand side

Since $\frac{1}{16}=(\frac{1}{2})^4$, the equation becomes $(\frac{1}{2})^4 = (\frac{1}{2})^{\frac{t}{5760}}$.

Step4: Solve for $t$

Since the bases are the same, we can set the exponents equal to each other: $4=\frac{t}{5760}$. Multiply both sides by 5760 to solve for $t$: $t = 4\times5760=23040$ years.

Answer:

23040 years