Answer:
t = 3,496x10⁹ years
Explanation:
The decay of ²³⁸U is:
²³⁸U → ²⁰⁶Pb + 8He + 6e⁻
Moles of ²⁰⁶Pb presents in 0,623mg are:
0,623x10⁻³g×(1mol / 206g) = 3,02x10⁻⁶ moles of ²⁰⁶Pb.
These moles are equals to moles of ²³⁸U before decay, that means, 3,02x10⁻⁶ moles²³⁸U
In grams:
3,02x10⁻⁶ moles²³⁸U× (238g / 1mol) = 7,20x10⁻⁴ g ²³⁸U = 0,720 mg²³⁸U
That means initial ²³⁸U was 1,000mg + 0,720mg =<em> 1,720mg</em>
Applying the formula:
ln (N₀/N) t₁₂ = t ln2
Where N₀ is initial amount of uranium (1,720mg), N is concentration of uranium (1,000mg), half-life time is a constant (t₁₂= 4,468x10⁹ years) and t is the time transcurred for the reaction. Replacing:
ln(1,720/1)*4,468x10⁹ years = t ln2
<em>t = 3,496x10⁹ years</em>
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I hope it helps!
