Question

A parchment fragment was discovered that had about 68% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment. (Round your answer to the nearest hundred years.)

Answer 1

Answer: The parchments is 3200 years old.

Explanation: The carbon-14 is radioactive with a half-life (t1/2) of 5730 years. This means that it will take 5730 years for an initial sample of the isotope is reduced to its half. Because of it, Carbon-14 can be used to determine the age of an object.

From the question, "68% as much 14C radioactivity as plant material today" means that there is still 68% of the initial sample of the isotope. To determine how many half-lives passed in that period of time, you use:

A = A₀·$\frac{1}{2^{n} }$ , in which

A is the amount left after a period of time;

A₀ is the initial amount;

n is the number of half-lives passed in the same period

It's known the remaining amount: A = $\frac{68}{100}$·A₀.

Substituing and calculating:

$\frac{68}{100}$·A₀ = A₀·$\frac{1}{2^{n} }$

0.68 = $\frac{1}{2^{n} }$

㏑0.68 = ㏑$(\frac{1}{2} )^{n}$

- 0.387 = n·( - 0.693)

n = $\frac{-0.387}{-0.693}$

n = 0.558

n is the amount of half-lives passed and it can be written as:

n = $\frac{t}{t_{\frac{1}{2} } }$ , where:

t is the period of time;

t1/2 is the half-life;

n = $\frac{t}{t_{\frac{1}{2} } }$

t = n · t1/2

t = 0.558 . 5730

t = 3200 years

The parchment is approximately 3200 years of age.