Question

A sample of uranium ore contains 6.73 mg of 238U and 3.22 mg of 206Pb. Assuming all of the lead arose from the decay of the uranium and that the half-life of 238U is 4.51 x 109years, determine the age of the ore

Answer 1

Answer:

The age of the ore is 4.796*10^9 years.

Explanation:

To solve this question, we use the formula;

A(t) =A(o)(1/2)^t/t1/2

where;

A(t) =3.22mg

A(o) = 6.731mg

t1/2 = 4.51*70^9 years

t = age of the ore

So,

A(t) =A(o)(1/2)^t/t1/2

3.22 = 6.73 (1/2)^t/4.51*10^9

Divide both sides by 6.73

3.22/6.73= (1/2)^t/4.51*10^9

0.47825= (0.5)^t/4.51*10^9

Log 0.4785 = t/4.51*10^9 • log 0.5

Log 0.4785/log 0.5 • 4.51*10^9 = t

t = 1.0634 * 4.51*10^9

t = 4.796*10^9

So therefore, the age of the ore is approximately 4.796*10^9 years.