Question
In a sample of rock, the ratio of $$^{206}Pb$$ to $$^{238}U$$ nuclei is found to be 0.5. The age in year of the rock is (given half - life of $${U^{238}}$$ is $$4.5 \times {10^9}$$ years)
A.
$$2.25 \times {10^9}$$
B.
$$4.5 \times {10^9}\ln 3$$
C.
$$4.5 \times {10^9}\frac{{\ln \left( {\frac{3}{2}} \right)}}{{\ln 2}}$$
D.
$$2.25 \times {10^9}\ln \left( {\frac{3}{2}} \right)$$
Answer :
$$4.5 \times {10^9}\frac{{\ln \left( {\frac{3}{2}} \right)}}{{\ln 2}}$$
Solution :
Suppose an initial radio nuclide $$I$$ decays to a final product $$F$$ with a half - life $${T_{\frac{1}{2}}}.$$
At any time, $${N_I} = {N_0}{e^{ - \lambda t}}$$
Number of product nuclei $$ = {N_F} = {N_0} - {N_I}$$
$$\eqalign{
& \frac{{{N_F}}}{{{N_I}}} = \frac{{{N_0} - {N_I}}}{{{N_I}}} = \left( {\frac{{{N_0}}}{{{N_I}}} - I} \right) \cr
& \frac{{{N_0}}}{{{N_I}}} = \left( {1 + \frac{{{N_F}}}{{{N_I}}}} \right) = 1 + 0.5 = 1.5 \cr
& \therefore \frac{{{T_{\frac{1}{2}}}\ln \left( {1.5} \right)}}{{\ln 2}} = 4.5 \times {10^9}\frac{{\ln \left( {\frac{3}{2}} \right)}}{{\ln 2}}{\text{year}} \cr} $$