Question

A rock outcrop was found to have 95.82% of its parent U- 238 isotope remaining. Approximate the age of the outcrop. The half-life of U- 238 is 4.5 billion years old.

A) 277 million years

B) 1.7 billion years

C) 14 million years

D) 21 million years

Answer 1

A) 277 million years

Explanation:

Here, decay of U-238 is first order reaction. (since most of the nuclear decays are first order).

so, the equation t =
`(1)/(k)` × ln(
`(A(0))/(A)`)

where, t = time from start of reaction

k = reaction constant

A(0) = initial concentration; A = present concentration.

⇒ 4.5 billion = 4500 million years =
`(1)/(k)`× ln2.

Now, given 95.82% of parent U-238 isotope is left.

⇒ t =
`(1)/(k)` × ln(
`(A(0))/(A)`)

t =
`(4500)/(ln2)` × ln(
`(100)/(95.82)`)

⇒ t ≅ 277 million years.