Question

While alive, an organism absorbs (radioactive) carbon 14 at such a rate that the proportion of carbon 14 in the organism remains at a known constant level. When it dies, it no longer absorbs carbon 14, so its proportion of carbon 14 begins to decrease. The half-life of carbon 14 is 5,600 years. A human skeleton is discovered whose level of carbon 14 is 15% that of a living human. To the nearest 100 years, how long ago did the person die

Answer 1

Answer:

The person has been dead for approximately 15,300 years

Explanation:

Available data:

• The half-life of carbon 14 is 5,600 years
• The human skeleton level of carbon 14 is 15% that of a living human

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

• C14 T₀ ⇒ Amount of carbon in a living body at time 0 = 100%
• C14T₁ ⇒ Amount of carbon in the dead body at time 1 = 15%
• λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
• T₀,₅ ⇒ The half-life of carbon 14 = 5600 years
• T₀ = 0
• T₁ = ???

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5600

λ = 0.000123

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (15%/100%) = Ln 0.15 = - 1.89

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 1.89 = - 0.000123 x T₁

T₁ = - 1.89 / - 0.000123

T₁ = 15,365 years

The person has been dead for approximately 15,300 years