In: Physics
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 200 Ci.
I got an answer of 30.99g but this was wrong.
the following soultion is for 290 Ci, if you didn't get any other response, kindly rate it.
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 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 = -A?e^(-?*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