Question

The contents of a rock have a 206Pb to 238U mass ratio of 0.135:1.00. Assuming that the rock did not contain any 206Pb at the time of its formation, determine the age of the rock. Uranium-238 decays to lead-206 with a half-life of 4.5 billion years. Express the time to two significant digits.

Answer 1

Answer: The age of the rock is $8.2\times 10^6yrs$

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 4.5 billion years = $4.5\times 10^9yrs$    

Putting values in above equation, we get:

$k=\frac{0.693}{4.5\times 10^9yrs}=1.54\times 10^{-8}yrs^{-1}$

We are given:

Mass ratio of 206-Pb to 238-U = 0.135 : 1.00

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant  = $1.54\times 10^{-8}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = [1.00 + 0.135] = 1.135 g

[A] = amount left after decay process = 1.00 grams

Putting values in above equation, we get:

$1.54\times 10^{-8}=\frac{2.303}{t}\log\frac{1.135}{1}\\\\t=8.2\times 10^6yrs$

Hence, the age of the rock is $8.2\times 10^6yrs$