The half-life for the radioactive decay of calcium-47 is 4.5 days.If a sample has an activity of 4.00uCi after 13.5 days, what was the initial activity of the sample?
13.5days/4.5 days = 3 half lives.
(Activity/23) = 4
Solve for activity.
GGHHH
Iron-59 has a half-life of 46 days. If the laboratory received a sample of 8.0 g of iron-59, the number of grams that are still active after 184 days is _
To find the initial activity of the sample, we can use the formula for radioactive decay, which relates the current activity A with the initial activity A₀ and the elapsed time t:
A = A₀ * (1/2)^(t / T)
Where A₀ is the initial activity, t is the elapsed time, and T is the half-life.
In this case, we are given that the half-life for calcium-47 is 4.5 days, the current activity is 4.00 microcuries (uCi), and the elapsed time is 13.5 days. We want to find the initial activity A₀.
Using the formula, we can substitute the given values:
4.00 uCi = A₀ * (1/2)^(13.5 / 4.5)
To solve for A₀, we need to isolate it. We can start by dividing both sides of the equation by (1/2)^(13.5 / 4.5):
4.00 uCi / (1/2)^(13.5 / 4.5) = A₀
Now let's simplify the expression inside the parentheses:
(1/2)^(13.5 / 4.5) = (1/2)^3 = 1/8 = 0.125
Substituting this value:
4.00 uCi / 0.125 = A₀
Dividing:
A₀ ≈ 32 uCi
Therefore, the initial activity of the sample was approximately 32 microcuries.