Iron-53 has a half-life of 8.51 minutes . After 15.0 minutes, 132 mg remain from the original sample.

Question

a) What was the original mass of the sample of iron- 53?
b) If only 58mg remain of the original sample, how much time has elapsed?

Answer 1

a) To find the original mass of the sample of iron-53, we can use the formula for radioactive decay:

mass remaining = original mass * (1/2)^(time elapsed / half-life)

Let's fill in the given information:
132 mg = original mass * (1/2)^(15.0 minutes / 8.51 minutes)

To solve for the original mass, we need to isolate it on one side of the equation. We can do this by dividing both sides by (1/2)^(15.0 minutes / 8.51 minutes):

132 mg / (1/2)^(15.0 minutes / 8.51 minutes) = original mass

Using a calculator, we find that the original mass is approximately 195.68 mg.

b) Similarly, we can use the same formula to find the time elapsed when only 58 mg remain:

58 mg = original mass * (1/2)^(time elapsed / 8.51 minutes)

To solve for the time elapsed, we need to isolate it on one side of the equation. We can do this by dividing both sides by original mass:

58 mg / original mass = (1/2)^(time elapsed / 8.51 minutes)

Taking the logarithm of both sides of the equation, we can simplify it:

log(58 mg / original mass) = (time elapsed / 8.51 minutes) * log(1/2)

time elapsed / 8.51 minutes = log(58 mg / original mass) / log(1/2)

time elapsed = (log(58 mg / original mass) / log(1/2)) * 8.51 minutes

Using the original mass calculated in part a) (approximately 195.68 mg), we can substitute it into the equation:

time elapsed = (log(58 mg / 195.68 mg) / log(1/2)) * 8.51 minutes

Using a calculator, we find that the time elapsed is approximately 4.43 minutes.