Question

A rock contains 0.275 mg of lead-206 for each milligram of uranium-238. The for the decay of uranium-238 to lead-206 is 4.5 x 10 9 yr. The rock was formed __________ yr ago.

Answer 1

Answer:

The rock is formed $1.7\times 10^9$ years ago.

Explanation:

Given that:

Half life = $4.5\times 10^9$ years

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

Where, k is rate constant

So,  

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

$k=\frac{ln\ 2}{4.5\times 10^9}\ years^{-1}$

The rate constant, k = $1.54\times 10^{-10}$ years⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the final concentration= 0.275 mg

$[A_0]$ is the initial concentration = 1 mg

Let $[A_t]$ = 1 mg

A rock contains 0.275 mg of lead-206 for each milligram of uranium-238. So,

$[A_0]$ = $1+\frac{238}{206}\times 0.275$ mg = 1.297 mg

Time = ?

So,  

$\frac{1}{1.297}=e^{-1.54\times 10^{-10}\times t}$

$\ln \left(\frac{1}{1.297}\right)=-1.54\times \:10^{-10}t$

$t=\frac{10^{10}\ln \left(1.297\right)}{1.54}$

$t=1.7\times 10^9$ years

The rock is formed $1.7\times 10^9$ years ago.