The age of rock that is containing 0.628 g Uranium-238 is 2032631864.35 years.
Half-life can be given as the time required by the object to reduced to half of its initial concentration. The concentration remained can be given as:
Final concentration = Initial concentration * 1/2 ^ (t/t1/2)
The initial concentration of rock has been the remaining uranium-238 and lead-206 cumulative concentration. Thus, the initial concentration is given as:
Initial concentration = Uranium-238 + Lead-206
Initial concentration = 0.068 g + 0.025 g
Initial concentration = 0.093g
The final concentration of Uranium-238 remained = 0.68 g
The half life given = 4.5 × 10⁹ years
The age (t) of the rock can be given as:
[tex]0.068=0.093\left(\frac{1}{2}\right)^{\frac{t}{4.5\times \:10^9}}\\\\\frac{t\times \:10^{-9}}{4.5}\ln \left(\frac{1}{2}\right)=\ln \left(\frac{68}{93}\right)\\\\t=2032631864.35961[/tex]
Thus, the age of rock that has remaining 0.068 g Uranium-238 is 2032631864.35 years.
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