A tumor is injected with 0.5 grams of Iodine-125, which has a decay
rate of 1.15% per day. To the nearest day, how long will it take for
half of the Iodine-125 to decay?
I get 99 days, the solution provided by my textbook says about 60 days.
Here's my working:
(Formula for the amount of Iodine remaining after $t$ days (where k is 1-1.15%))
$$A=A_oe^{\frac{ln0.5}{98.85}t}$$
The patient was injected with 0.5 grams so I want to know how long it takes for half of that, 1/4 grams to remain:
$$\frac{1}{4}=\frac{1}{2}e^{\frac{ln0.5}{98.85}t}$$
$$1=2e^{\frac{ln0.5}{98.85}t}$$
$$\frac{1}{2}=e^{\frac{ln0.5}{98.85}t}$$
$$ln\frac{1}{2}=\frac{ln0.5}{98.85}t$$
$$t=\frac{ln\frac{1}{2}}{\frac{ln\frac{1}{2}}{98.85}}=98.85$$
Where did I go wrong and how can I arrive at 60 days?
Best Answer
Your equation $A = A_0 e^{\frac{\ln 0.5}{98.85}t}$ is incorrect.
If we start with the general model, $A(t)=A_0e^{kt}$, and use the fact that after one day, $.9885$ of the initial amount remains, we get $$A_0e^k=.9885A_0$$ So $k=\ln(.9885)$
The model becomes $A(t)=A_0e^{\ln(.9885)\cdot t}$
From there you should be able to get the half-life.