Question

One of the many isotopes used in cancer treatment is gold ¹⁹⁸₇₉Au, with a half-life of 2.70 days. Determine the mass of this isotope that is required for an activity of 213 Ci.

Answer 1

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

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

ln(2) / ? = 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)
? = 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 213 Ci (which is also 7.881x10^12 decays per second). For this equation t = 0 because we're talking about the initial activity.
-7.881x10^12 = -A?e^(-?*0)
(the left side of the equation is negative because we're losing isotopes and not gaining them)
Solve for A
A = 2.65x10^18 isotopes
Now let's find how much mass is in this many isotopes.
(2.65x10^18)*(198u) =
5.25x10^20 u
Now convert to mg to get your final answer
8.71 mg