Question

2. Before the year 1900 the activity per unit
mass of atmospheric carbon due to the
presence of 14C averaged about 0.255 Bq
per gram of carbon. (a) What fraction
of carbon atoms were 14C? (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?
Half-life of 14C is 5730 years?
3y1012​

Answer 1

Solution :

Given data :

$T_{1/2} = 5730\ y$

∴  $\lambda =\frac{0.693}{5730 \times 3.156 \times 10^7} \ s^{-1}$

       $=3.832 \times 10^{-12} \ s^{-1}$

A = 0.255 Bq per gram of carbon in part (a), M = 500 mg = $500 \times 10^{-3} \ g$

174 days in one hour = $\frac{174}{3600}$ dis/s = 0.04833 dis/s in part (b) (per 500 mg)

(a) $A=N \lambda$

    ∴ $N =\frac{A}{\lambda}$

           $=\frac{0.255}{3.832 \times 10^{-12}}$

          $=6.654 \times 10^{10}$

Number of atoms in 1 g of carbon $=\frac{6.02 \times 10^{23}}{12}=5.017 \times 10^{23}$

$\frac{5.017 \times 10^{22}}{6.654 \times 10^{10}} = 0.7539 \times 10^{12}$

So, one $^{14}C$ atom per $0.7539 \times 10^{12}$ atoms of carbon

∴ Four $^{14}C$ atom per $3 \times 10^{12}$ atoms of carbon

(b). Present activity per gram $=\frac{0.04833}{500 \times 10^{-3}}$

                                                = 0.09666 dis/s per gram

$A_0=0.255$ dis/s per gram.

Now,

$A(t) = A_0e^{-\lambda t}$

$\lambda t = 2.303 \ \frac{\log_{10}A_0}{A}$

$\lambda t=2.303 \log_{10}\left(\frac{0.255}{0.09666}\right)$

$t=\frac{2.303 \ \log 2.638}{3.832 \times 10^{-12}}$

$t=\frac{2.303 \times 0.4213}{3.832 \times 10^{-12}}$

$t=25.31 \times 10^{10} \ s$

$t=\frac{25.32 \times 10^{10}}{3.156 \times 10^{7}}$

t = 8023 years

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