Question

If a substance is radioactive, this means that the nucleus is unstable and will therefore decay by any number of processes (alpha decay, beta decay, etc.). The decay of radioactive elements follows first-order kinetics. Therefore, the rate of decay can be described by the same integrated rate equations and half-life equations that are used to describe the rate of first-order chemical reactions:

lnAtA0=−kt

and

t1/2=0.693k

where A0 is the initial amount or activity, At is the amount or activity at time t, and k is the rate constant.

By manipulation of these equations (substituting 0.693/t1/2for k in the integrated rate equation), we can arrive at the following formula:

fraction remaining=AtA0=(0.5)n

where n is the number of half-lives. The equation relating the number of half-lives to time t is

n=tt1/2

where t1/2 is the length of one half-life

1.Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 35.0 % of an Am-241 sample to decay?

2.A fossil was analyzed and determined to have a carbon-14 level that is 80 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

Answer 1

Given

t1/2 = 432 years

if 35 % sample decay then 100 - 35 = 65 % of sample will be remaining

fraction remaining = 0.65 = At / Ao = (0.5)ⁿ where n = t/t_1/2

taking ln on both sides

ln 0.65 = n ln 0.5

n = ln 0.65 / ln 0.5 = 0.62

n = t/t_1/2 = 0.62

t = 0.62 * t_1/2 = 0.62 * 432 = 268.5 years

so it will take 268.5 years for 35 % of the sample to decay

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2)

given

t_1/2 = 5730 years

C 14 remaining is 80 % of orginal

fraction remaining = 0.8 = At / Ao = (0.5)ⁿ where n = t/t_1/2

taking ln on both sides

ln 0.8 = n ln 0.5

n = ln 0.8 / ln 0.5 = 0.322

n = t/t_1/2 = 0.322

t = 0.322 * t_1/2 = 0.322 * 5730 = 1844.65 years

so fossil is 1844.65 years old