The half-life of iodine -131 is 16 days. If you start with 100 gram sample, how many half-lives have past when you have less than a gram left?
To determine the number of half-lives that have passed when you have less than a gram left, we can use the formula:
N = N0 * (1/2)^(t / T)
Where:
N = final amount of the substance (in this case, less than a gram)
N0 = initial amount of the substance (100 grams)
t = time that has passed (the number of half-lives)
T = half-life of the substance (16 days)
Let's solve the equation for t:
N = N0 * (1/2)^(t / T)
Rearranging the equation, we have:
(1/2)^(t / T) = N / N0
Taking the logarithm of both sides (base 0.5):
log2 [(1/2)^(t / T)] = log2 (N / N0)
Using the logarithmic property, we can move the exponent down as a coefficient:
(t / T) * log2 (1/2) = log2 (N / N0)
Recognizing that log2 (1/2) = -1, the equation simplifies to:
(t / T) * -1 = log2 (N / N0)
Rearranging the equation further:
t / T = -log2 (N / N0)
Now, we can substitute the given values:
t / 16 = -log2 (less than 1 / 100)
Since the condition is when there's less than a gram left, N is less than 1 gram. Let's approximate it to 0.99 grams:
t / 16 = -log2 (0.99 / 100)
Solving the equation for t:
t = -16 * log2 (0.99 / 100)
Using a calculator, we find:
t ≈ -16 * (-0.014434)
t ≈ 0.231 milliseconds
So, approximately 0.231 milliseconds will pass when you have less than a gram left. Since this is a very short time, we can conclude that no half-lives have passed.
2n = No/N.
No = 100
N = 1. Solve for n.
I get n - 6.64 half lives which means you need 7 half lives to get to less than 1 g. Six half lives leaves 1.56 g.