Question

One of the many isotopes used in cancer treatment is 79¹⁹⁸Au, with a half-life of 2.70 d. Determine the mass of this isotope that is required to give an activity of 175 Ci.

Answer 1

Radioactive decay equation:
N = number of atoms left in Au
N₀ = initial number of atoms in Au
A = decay constant

A₀ = intial decay constant

N = N₀e^-λt

By deriving that previous equation you can find the rate of decay equation:
A= rate of decay in isotopes per second
A = -A₀ λe^-λt

λ = 0.693/T_1/2

1 Ci is 3.7x10^10 decays per second

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

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 atoms needed to have an activity of 175 Ci (which is also 1.073x10^13 decays per second).

For this equation t = 0 because we're talking about the initial activity.
-6.475x10¹² = -Aλe^-λ*0
(the left side of the equation is negative because we're losing isotopes and not gaining them)
Solve for A
A = 2.179x10¹⁸ atoms
Now let's find how much mass is in this many isotopes.
(2.179x10¹⁸)X(198u) =4.31X10²⁰ u
Now convert to mg to get your final answer
7.16X10^-7g