Question

How old is a bone in which the Carbon-14 in it has undergone 3 half-lives? A) 11,400 years old Eliminate B) 17,100 years old C) 34,200 years old D) 57,000 years old

Answer 1

The age of a bone that has undergone 3 half-lives of Carbon-14 is 17,100 years, calculated by multiplying the half-life of Carbon-14 (5,700 years) by three.

Step-by-step explanation:

To determine the age of a bone in which Carbon-14 has undergone 3 half-lives, we use the information that the half-life of C-14 is approximately 5,700 years. After one half-life, half of the original C-14 would remain; after two half-lives, a quarter would remain; and after three half-lives, an eighth would remain.

To calculate the total number of years for three half-lives, we would multiply the half-life (5,700 years) by three. Therefore, 5,700 years × 3 equals 17,100 years.

The bone is 17,100 years old, which corresponds to Option B).

Answer 2

The answer is B 17,100,

(1,250 is the third half life of carbon 14)

Step-by-step explanation:

The carbon 14 remaining in 0 years is 10,000. The carbon 14 remaining after its first half life would be 5,000, which is 5,700 years after death, the second half live would be 2,500 which is 11,400 years after death, the third half life is 1,250 which is 17,100 years after death. (divide the full life 10,000 by two)

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