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Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of ^{14}C averaged about 0.255 Bq per gram of carbon.

(a) What fraction of carbon atoms were ^{14}C?

(b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? The half-life of ^{14}C is 5730 years.

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#### Solution

**Data: **T_{1/2} = 5730 y

∴ `lambda = 0.693/(5730 xx 3.156 xx 10^7)"s"^-1`

= 3.832 × 10^{-12} s^{-1}, A = 0.255 Bq per gram of carbon in part (a); M = 500 mg = 500 × 10^{-3} g,

174 decays in one hour = `174/3600` dis/s = 0.04833 dis/s in part (b) [per 500 mg]

(a) A = Nλ

∴ N = `"A"/lambda = 0.255/(3.832 xx 10^-12)`

= 6.654 × 10^{10}

Number of atoms in 1 g of carbon = `(6.02 xx 10^23)/12 = 5.017 xx 10^22`

`(5.017 xx 10^22)/(6.654 xx 10^10) = 0.7539 xx 10^12`

∴ 1 ^{14}C atom per 0.7539 x 10^{12} atoms of carbon

∴ 4 ^{14}C atoms per 3 x 10^{12} atoms of carbon

(b) Present activity per gram = `0.04833/(500 xx 10^-3)`

= 0.09666 dis/s per gram

A_{0} = 0.255 dis/s per gram

Now, A(t) = `"A"_0"e"^(-lambda"t")`

∴ `lambda"t" = 2.303 log_10 "A"_0/"A" = 2.303 log_10 (0.255/0.09666)`

∴ t = `(2.303 log 2.638)/(3.832 xx 10^-12) = ((2.303)(0.4213))/(3.832 xx 10^-12)`

= 25.31 × 10^{10} s

`= (25.32 xx 10^10)/(3.156 xx 10^7) = 8023` years

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