Question

Compared with the fibers of cotton plants growing today, what is the relative ratio radioactivity in the old material vs the relative amount radioactivity in the new material? The half-life of 14C is 5730 y. (A0 = original radioactivity and At = current radioactivity).
For a quipu sample that is 4800 years old, we will calculate the ratio of 14C/12C in the sample.
At/A0 = ?

Answer 1

Answer:

0.56

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/At)

t1/2 = half life of the C-14 = 5730 y

t = time elapsed = 4800 y

At = Activity of C-14 at time t

Ao= Activity of a living C-14 sample

0.693/5730 = 2.303/4800 log (Ao/At)

1.2 * 10^-4 = 4.8 * 10^-4 log (Ao/At)

log (Ao/At) = 1.2 * 10^-4/4.8 * 10^-4

log (Ao/At) = 0.25

Ao/At = Antilog (0.25)

Ao/At = 1.778

Hence;

At/Ao = (1.778)^-1

At/Ao = 0.56

Answer 2

The answer is 0.56, for the cotton that is 4800 years old and has a similar quality like those of today.

What is the sample of the ratio of cotton ?

The plant has grown in today in a relative ratio of radioactivity in the old matter vs the relative amount of radioactivity in the new matter. The half-life for which in given as 14C is 5730 y.

As per the formula 0.693/t1/2 = 2.303/t log (Ao/At)

t1/2 = half life of the C-14 = 5730 y, t = time elapsed = 4800 y

At = Activity of C-14 at time t Ao= Activity of a living C-14 sample

0.693/5730 = 2.303/4800 log

(Ao/At) 1.2 * 10^-4 = 4.8 * 10^-4 log (Ao/At)

Thus At/Ao = 0.56.

Find out more information about the fibers of cotton.

brainly.com/question/13811890.