Answer:
The answer is below
Explanation:
Let m be the mass of each of lead-206 and uranium-238 found in the rock. The radioactive decay is given as:
[tex]U_t=U_oe^{-\lambda t}\\\\U_t=number\ of\ undecayed\ U238\ after\ time \t,\\U_o=number\ of\ undecayed\ U238\ initially,\lambda=decay\ constant\\\\But\ U_o=U_t+Pb_t\\\\U_t=U_t+Pb_t(e^{-\lambda t})\\\\e^{-\lambda t}=\frac{U_t}{U_t+Pb_t} \\\\e^{\lambda t}=\frac{U_t+Pb_t}{U_t} \\\\e^{\lambda t}=1+\frac{Pb_t}{U_t}[/tex]
half life of uranium-238 = 4.468 * 10⁹ years
λ = 0.693 / half life = 0.693 / 4.468 * 10⁹ years = 0.1551 * 10⁻⁹
[tex]n_{Pb_t}=\frac{m}{206} \\\\n_{U_t}=\frac{m}{238}\\\\1+\frac{\frac{m}{206}}{\frac{m}{238}}=e^{\lambda t} \\\\e^{\lambda t} =2.1553\\\\ln(e^{\lambda t} )=ln(2.1553)\\\\\lambda t=0.7679\\\\t=\frac{0.7679}{\lambda} \\\\t=\frac{0.7679}{0.1551*10^{-9}} \\\\t=3.43*10^9\ years[/tex]
