Question

Radioactive Decay of Carbon 14

Years since death C14 atoms remaining per 1.0 x 108 C12 atoms
0 10,000
5,700 5,000
11,400 2,500
17,100 1,250
22,800 625
28,500 312
34,200 156
39,900 78
45,600 30
51,300 20
57,000 10
62,700 5
68,400 2

How old is a bone in which the Carbon-14 in it has undergone 3 half-lives?

A. 11,400

B.17,100

C. 34,200

D. 57,000

Please hurry I have a test

Answer 1

:Answer

1.0/5

7

MissPhiladelphia Ambitious

17.2K answers

145.1M people helped

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 = A