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·
`(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 =
`(68)/(100)`·A₀.

Substituing and calculating:

`(68)/(100)`·A₀ = A₀·
`(1)/(2^(n) )`

0.68 =
`(1)/(2^(n) )`

㏑0.68 = ㏑
`((1)/(2) )^(n)`

- 0.387 = n·( - 0.693)

n =
`(-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_{(1)/(2) } }` , where:

t is the period of time;

t1/2 is the half-life;

n =
`\frac{t}{t_{(1)/(2) } }`

t = n · t1/2

t = 0.558 . 5730

t = 3200 years

The parchment is approximately 3200 years of age.