Answer:
C-14 when formed naturally decays by beta emission to N-14. Therefore, there's a 100% probability of naturally decaying from its birth past 6000 yrs and on through 11,200 yrs and on to an infinity of timed half-lives.
Explanation:
The question is somewhat ambiguous in its wording but the following note will focus on defining the 1st order decay trend of C-14 and the half-life values that can be determined from the 1st order decay equation.
Radioactive decay is described by 1st order kinetics and follows an exponential trend shown graphically below. Such a trend is represented by the expression A = A₀·exp(-k·t) where A = final activity, A₀ = initial activity, k = rate constant and t = time of decay.
Also in describing radioactive decay process is 'half-life'. It is defined as the time needed for the original amount of material to decrease to 1/2 of its original amount. The 'half-life' equation is derived from the above 1st order decay equation and is given by the expression t(1/2) = 0.693/k. This equation allows one to determine the time needed for 'one' half-life. The same equation can be used to determine the 'second' half-life, or the time it takes for the original material to decay to 1/4th of the original amount and so on.
In consideration of the decay of C-14 which has a published 1st half-life of 5,600 years, then a second half-life equals 2(5600 yrs) or 11,200 yrs and a 3rd half-life would be 3(5600 yrs) or 16,800 yrs and so on.
So, since C-14 naturally decays by beta emission which is high energy electron emissions (β⁻) to give N-14. Such will continue to decay past 5,600 years on to a 2nd half-life for as many half-lives one wished to consider.
