42.5k views
5 votes
If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 3.2 to 0.4 mg in 39.6 h, what is the half-life of iodine-123?

User Kauppfbi
by
7.5k points

1 Answer

3 votes

Answer:

Half life = 13.197 hour

Step-by-step explanation:

Given:

Old amount (A₀) = 3.2

New amount (A) = 0.4

Radiation decay time (t) = 39.6 hour

Half life = T(1/2)

Find:

Half life = T(1/2) = T

Computation:

A = A₀
e^{-((0.693t)/(T) )}


e^{-((0.693t)/(T) )} = 0.4 / 3.2

-[27.4428 / T] = In (0.125)

-[27.4428 / T] = -2.0794

[27.4428 / T] = 2.0794

T = 13.197

Half life = 13.197 hour

User Tikinoa
by
6.6k points