In this question, we have to calculate the age of a fossil after the decay of Carbon-14 436 times, with a half-life of 5730 years, and in order to do that, we can multiply both values, one example to understand this step better is:
Let's say we have a 2 times decay of a certain amount of Carbon-14, we can say that 11,460 years have passed by, and we can check it:
x grams of C 14 = 0 years
x/2 grams of C 14 = 5730 years, 1st decay
x/4 grams of C 14 = 11460 years, 2nd decay
Therefore with 436 decays
5730 * 436 = 2,498,280 years this is the age of this fossil
Part B:
For this part, we will need to use the Half-Life formula, which is the following:
N(t) = N0(1/2)^t/t1/2
Where:
N(t) = quantity of the substance remaining, 2.2*10^-78
N0 = initial quantity of the substance
t = time elapsed, 2,498,280 years
t1/2 = half life of the substance, 5730 years
Now we add these values into the formula:
2.2*10^-78 = N0(1/2)2,498,280/5730
2.2*10^-78 = N0(1/2)^436
2.2*10^-78 = 5.63*10^-132N0
N0 = 2.2*10^-78/5.63*10^-132
N0 = 3.91*10^53 was the initial quantity
