To solve the given problems, we'll use the formula for exponential decay:
N(t) = N0 * (1/2)^(t/h)
Where:
N(t) is the amount remaining after time t
N0 is the initial amount
t is the elapsed time
h is the half-life
a. How much will remain after 69 years?
Using the formula, we have:
N(t) = N0 * (1/2)^(t/h)
N(69) = 33 * (1/2)^(69/12.4)
N(69) ≈ 33 * (1/2)^5.5645
N(69) ≈ 33 * 0.097
N(69) ≈ 3.201 grams
Approximately 3.201 grams will remain after 69 years.
b. How long until there is 5 grams remaining?
Using the formula, we need to solve for t:
5 = 33 * (1/2)^(t/12.4)
Divide both sides by 33:
(1/6.6) = (1/2)^(t/12.4)
Taking the logarithm base 2 of both sides:
log2(1/6.6) = (t/12.4) * log2(1/2)
log2(1/6.6) = (t/12.4) * (-1)
Rearranging the equation:
(t/12.4) = log2(1/6.6)
Multiplying both sides by 12.4:
t = 12.4 * log2(1/6.6)
Using a calculator, we find:
t ≈ 33.12 years
Approximately 33.12 years are required until there is 5 grams remaining.
c. How much of an initial sample would you need to have 50 grams remaining in 22 years?
Using the formula, we need to solve for N0:
50 = N0 * (1/2)^(22/12.4)
Divide both sides by (1/2)^(22/12.4):
50 / (1/2)^(22/12.4) = N0
Using a calculator, we find:
N0 ≈ 74.91 grams
To have approximately 50 grams remaining in 22 years, the initial sample would need to be approximately 74.91 grams.