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

Answer:

A) 277 million years

Step-by-step explanation:

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

so, the equation t = $\frac{1}{k}$ × ln($\frac{A(0)}{A}$)

where, t = time from start of reaction

           k = reaction constant

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

     ⇒ 4.5 billion = 4500 million years = $\frac{1}{k}$× ln2.

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

⇒ t = $\frac{1}{k}$ × ln($\frac{A(0)}{A}$)

  t = $\frac{4500}{ln2}$ × ln($\frac{100}{95.82}$)

⇒ t ≅ 277 million years.