A tumor is injected with 0.6 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

The decay rate of iodine-125 is 1.15% per day, which means that after one day, 1.15% of the iodine-125 will decay.

Let's use the formula for exponential decay:

N(t) = N0 * e^(-λt)

where N(t) is the amount of iodine-125 remaining after time t, N0 is the initial amount of iodine-125 (0.6 grams), λ is the decay constant (ln(2)/half-life), and t is the time elapsed.

We want to find the time it takes for half of the iodine-125 to decay, which means that N(t) = 0.5*N0.

0.5*N0 = N0 * e^(-λt)

0.5 = e^(-λt)

Taking the natural logarithm of both sides:

ln(0.5) = -λt

t = -ln(0.5)/λ

The half-life of iodine-125 is 60.14 days, which means that λ = ln(2)/60.14 = 0.01154 per day.

Plugging this into the equation:

t = -ln(0.5)/0.01154 ≈ 60 days

Therefore, it will take approximately 60 days for half of the iodine-125 to decay.

Answer 2

1.15% decline = 98.85% remaining
N(t) = 0.6 * 0.985^t
so solve for t in
0.6 * 0.985^t = 0.3
0.985^t = 0.5
t log0.985 = log0.3
t = 79.66