Question

. The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood? (log 3= 0.4771, log 7 = 0.8540 , Half-life of C-14 = 5730 years )

Answer 1

k = 0.693/t1/2

k = 0.693/5730 years-1

t = 2.303 log Co

k Ct

let Co = 1 Ct = 3/10 so Co/Ct = 1/ (3/10) = 10/3

t = 2.303 x 5730 log 10

0.693 3

t = 19042 x (1-0.4771) = 9957 years

Answer 2

The C-14 content of an ancient piece of wood was found to have three-tenths of that in living trees. The ancient piece of wood is approximately 14,501 years old.

• The C-14 content of an ancient piece of wood is three-tenths of that in living trees, which means the ratio of C-14 in the ancient wood to that in living trees is 3/10.

• The C-14 content of living trees is assumed to be constant and is used as a reference point to determine the age of ancient objects. The C-14 content of living trees is considered to be 100%, so the ratio of C-14 in the ancient wood to that in living trees can be written as 3/100.

• The half-life of C-14 is 5730 years. This means that after 5730 years, the C-14 content of a sample will have decreased to half of its original value.

• If the ratio of C-14 in the ancient wood to that in living trees is 3/100, this means that the C-14 content of the ancient wood has decreased by a factor of 100/3 = 33.3 times.

• To determine how many half-lives the C-14 content of the ancient wood has decreased by a factor of 33.3, we can use the formula:

• t = (T/H) * ln(R)

• where t is the age of the object, T is the half-life of the isotope, H is the initial C-14 content (100% in this case), and R is the ratio of C-14 in the ancient wood to that in living trees (3/100 in this case).

• Plugging these values into the formula, we get:

• t = (5730 years / 100%) * ln(3/100)

         = 5730 years * (-2.52)

         = -14,501 years

Therefore, the ancient piece of wood is approximately 14,501 years old.

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