Answer:
The age of the Igneous Sill is 1.456 × 10¹⁰ years
Explanation:
The radioactive decay is given as follows;
[tex]U_t = U_0 e^{-\lambda t}[/tex]
Also, where 1 Uranium 238 atom disintegrates to produce 1 Lead 206 atom, we write;
[tex]U_t = (U_t + Pb_t) e^{-\lambda t}[/tex]
Such that;
[tex]\frac{Pb_t}{U_t} = e^{-\lambda t} - 1[/tex]
The decay constant, λ is given by the following;
[tex]\lambda = \frac{0.693}{4.468 \times 10^9} = 1.55 \times 10^{-10} \ year^{-1}[/tex]
Percentage by mass of Uranium = 91.7%
Percentage by mass of Uranium = 8.3%
Number of moles of Uranium 238 and Lead 206 present is therefore;
[tex]nPb_t = \frac{8.3}{205.974 465} = 0.0403[/tex]
[tex]nU_t = \frac{91.7}{238.05078826} = 0.385[/tex]
Therefore we have;
[tex]\frac{0.0403}{0.385} = e^{-\lambda t} - 1 = e^{-1.55 \times 10^{-10} \times t} - 1= 0.1046[/tex]
From which t is found to be t = 14555074285.2 = 1.456 × 10¹⁰ years.
Answer:
[tex]t = 558791705.6\,years[/tex]
Explanation:
After reading the statement of the problem, part of the initial mass of Uranium-238 has become into Lead-206. The decayment of the isotope is modelled by the following expression:
[tex]\frac{m}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The Uranium-238 has a half-life of 4.47 billion years and the time constant as function of the half-life is:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]
[tex]\tau = \frac{4.47\times 10^{9}\,years}{\ln 2}[/tex]
[tex]\tau = 6.449\times 10^{9}\,years[/tex]
The age of the Igneous Sill is found by clearing the variable in the decayment equation:
[tex]\ln \frac{m}{m_{o}} = -\frac{t}{\tau}[/tex]
[tex]t = - \tau \cdot \ln \frac{m}{m_{o}}[/tex]
[tex]t = - (6.449\times 10^{9}\,years)\cdot \ln 0.917[/tex]
[tex]t = 558791705.6\,years[/tex]
