JEE Advanced Chemistry: Nuclear Chemistry questions with solutions

Q1. The C-14 content in a fossil is presently 1/5 times that found in the present atmosphere. After the death of this organism, a nuclear explosion caused a 25% increase in atmospheric C-14 content. If the age of the fossil is x * 10⁴ years, find x. Given: half-life of C-14 = 6000 years.

1. 1.20
2. 1.6
3. 2.4
4. 1.8

Answer: 1.20

At time of death, atmospheric C-14 = N0 (original). After death, nuclear explosion raised it by 25%, so current atmospheric = 1.25*N0. Current fossil C-14 = (1/5)*1.25*N0 = N0/4 (relative to original N0 at death). Since N/N0 = 1/4 = (1/2)², exactly 2 half-lives have passed. Age = 2 * 6000 = 12000 years = 1.2 * 10⁴ years.