A lab has 40 mg of radioactive iodine. After 24 days there are only 5 mg remaining. What is
the half-life of the iodine?
To find the half-life of the iodine, we can use the half-life formula:
N(t) = N(0) * (1/2)^(t/h)
Where:
N(t) = remaining amount of iodine after t time (5 mg)
N(0) = initial amount of iodine (40 mg)
t = time passed (24 days)
h = half-life of iodine (unknown)
Plugging in the given values:
5 mg = 40 mg * (1/2)^(24/h)
To solve for h, we can take the logarithm base 1/2 of both sides:
log base (1/2) (5/40) = 24/h
Simplifying the equation:
log base (1/2) (1/8) = 24/h
To convert the logarithm with a base of 1/2 to a more common base, we can use the change of base formula:
log base (1/2) (1/8) = log(1/8) / log(1/2)
Using a calculator, we can find that:
log base (1/2) (1/8) ≈ -3
So, the equation becomes:
-3 = 24/h
To solve for h, we isolate it:
h = 24 / -3
h = -8
The half-life of the iodine is approximately 8 days.