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

For this problem, we use the formula for radioactive decay which is expressed as follows:

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant. 

We calculate as follows:

at hal-life equal to 5700,

An = Aoe^-kt
0.5 = e^-k(5700)
k = 1.22x10^-4

An = Aoe^-kt
1.5 = e^-1.22x10^-4(t)
t = 3323 years

Answer 2

Answer:

The answer is B 17,100,

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

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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