​​$$\text{In a sample of charcoal, the remaining fraction of } ^{14}\text{C is } 2 \times 10^{-2}. \text{ Its age (in years) is closest to} \\\text{[Given } t_{1/2} \text{ of } ^{14}\text{C} = 5730 \text{ years]}$$​​

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Correct option is A

​During its life, a plant or animal is in equilibrium with its surroundings by exchanging carbon either with the atmosphere or through its diet. It will, therefore, have the same proportion of ¹⁴C as the atmosphere, or in the case of marine animals or plants, with the ocean. Once it dies, it ceases to acquire ¹⁴C, but the ¹⁴C within its biological material at that time will continue to decay, and so the ratio of ¹⁴C to ¹²C in its remains will gradually decrease. Because ¹⁴C decays at a known rate, the proportion of radiocarbon can be used to determine how long it has been since a given sample stopped exchanging carbon – the older the sample, the less ¹⁴C will be left.

The equation governing the decay of a radioactive isotope is:

$$N = N_0 e^{-\lambda t}$$

​where N₀ is the number of atoms of the isotope in the original sample (at time t = 0, when the organism from which the sample was taken died), and N is the number of atoms left after time t. λ is a constant that depends on the particular isotope; for a given isotope it is equal to the reciprocal of the mean-life – i.e. the average or expected time a given atom will survive before undergoing radioactive decay.

t_1/2= 0.693/λ (t_1/2 is the half life)​

Ans. 3.2×10⁴​