Arock contains 0.623 mg of 206pb for every 1.000 mg of 238u present. assuming that no lead was originally present, that all the 206pb formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between 238u and 206pb is negligible, calculate the age of the rock.

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 = 1,720mg

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

t = 3,496x10⁹ years

I hope it helps!

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.