Question

A tumor is injected with 0.8 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?

Answer 1

If the initial amount is 0.8 grams, half of this amount would be 0.4 grams. But first, let's construct an equation for the decay rate. That would be:

Amount left = Initial amount*(1 - Decay Rate)*Time
0.4= 0.8(1 - 0.0115)(t)
Solving for t,
t = 0.51 days

Technically, it's just half day. But if we convert to the nearest day, the answer would be 1 day.

Answer 2

Based on the radioactive decay model:

A = A₀exp(-kt) ----(1)

where

A = concentration of the radioisotope at time t

A₀ = initial concentration

k = decay constant

t = time

Now, based on the given data:

A₀ of I-125 = 0.8 g

Since the decay rate is 1.15% per day, the amount of I-125 left after t = 1 day is:   A = 0.8 - (1.15/100)*0.8 g = 0.7908 g

substituting for A, A₀ and t in eq(1) we get:

0.7908 = 0.8 exp(-k*1)

k = 0.01157 day-1

The decay constant is related to t1/2 as follows:

t1/2 = 0.693/k

= 0.693/0.01157 = 59.89 days

Ans: The t1/2 for I-125 approximately 60 days