Answer:
Explanation:
Here we will be using the radiactive decay equations :
Nt/N₀ = e^-kt where k is the decay constant per year
t= time in years
Nt = amount of ²³⁸ U present at time t in μg ( μ =microgram)
N₀ = amount of ²³⁸ U originally present in μg
We notice we dont have k, but this value can be obtained from t₁/₂ through
k = 0.693 / t₁/₂
which is derived from Nt/N₀ = e^-kt for the half-life where Nt/N₀ = 0.5
The hints given in the problem helps us greatly to determine N and N₀
All the ²⁰⁶ Pb comes from ²³⁸ U and we know its mass = 0.730 in μg, so lets find moles ²³⁸ U
0.730 ug x 1 mol ²⁰⁶ Pb/206 umol = 0.3544 x 10⁻³ umol
This means 0.3544 x 10⁻³ in μmol of ²³⁸ U decayed
To find the mass decayed multiply by the atomic weight:
0.3544 x 10⁻³ in μmol x 238 g/ m in μmol = 8.434 x 10⁻² in μg ²³⁸ U decayed
N₀ = 8.434 x 10⁻² in μg + 1.000 in μg = 1.843 in μg
Now lets calculate the value for k and we will finally can compute the age of the sample:
k = 0.693/ 4.47 x 10⁹ yrs⁻¹ =1.550 x 10⁻¹⁰ yrs⁻¹
Nt/N₀ = e^-kt ⇒ (1.000 in μg/ 1.843 μg) = e ^- ( 1.550 x 10⁻¹⁰yrs⁻¹ / x t)
Taking natural log to both sides of the equation:
ln (1.000/1.843 ) = - 1.550 x 10⁻¹⁰ yrs⁻¹ x t
t= 3.94 x 10⁹ yrs = 3.94 billion years
To check our answer note that the time calculated is less than a half-life and the mass decayed is less than half the mass originally present.
(Note I left the mass in units of micrograms but could have also worked in grams but it would have been messier.)
