Question

A sample of U²³⁸ (half-life = 4.5 × 10⁹ years) ore is found to contain 23.8 g of U²³⁸ and 20.6 g of Pb²⁰⁶. The age of the ore is ______ × 10⁹ years.

Options

• 2.40

• 3.50

• 4.50

• 5.50

Answer 1

A sample of U²³⁸ (half-life = 4.5 × 10⁹ years) ore is found to contain 23.8 g of U²³⁸ and 20.6 g of Pb²⁰⁶. The age of the ore is 4.50 × 10⁹ years.

Explanation:

\[\ce{_92U^238 -> _82Pb^206 + 8 _2He^4 + 6 _{-1}e^0}\]

Pb present = `20.6/206` = 0.1 g atom = U decayed

U present = `23.8/238` = 0.1 g atom

Thus, N = 0.1 g atom

N₀ U present + U decayed = 0.1 + 0.1 = 0.2 g atom

Now t = `(2.303 xx 4.5 xx 10^9)/0.693 log_10  0.2/0.1`

t = 4.5 × 10⁹ year