Question

The isotope 198 79 Au (atomic mass = 197.968 u) of gold has a half-life of 2.69 days and is used in cancer therapy. What mass (in grams) of this isotope is required to produce an activity of 308 Ci?

Answer 1

Here's the stuff you need to know:

Radioactive decay equation:
y = number of isotopes left
A = initial number of isotopes
λ = decay constant
y(t) = Ae^(-λt)

By deriving that previous equation you can find the rate of decay equation:
r = rate of decay in isotopes per second
r(t) = -Aλe^(-λt)

ln(2) / λ = T(1/2)

1 Ci is 3.7x10^10 decays per second

1 mass unit (u) = 1.66x10^-27 kg


Ok first off let's convert the half life of 2.70 days into seconds and you get a half life of:
T(1/2) = 233300 sec
Now find the decay constant by dividing the half life from ln(2)
λ = 2.971x10^-6
Now that we have the decay constant, we can use the rate of decay equation to solve for the number of isotopes needed to have an activity of 290 Ci (which is also 1.073x10^13 decays per second). For this equation t = 0 because we're talking about the initial activity.
-1.073x10^13 = -Aλe^(-λ*0)
(the left side of the equation is negative because we're losing isotopes and not gaining them)
Solve for A
A = 3.611x10^18 isotopes
Now let's find how much mass is in this many isotopes.
(3.611x10^18)*(198u) =
7.150x10^20 u
Now convert to mg to get your final answer
1.187 mg