Iron-53 has a half-life of 8.51 minutes. After 15.0 minutes, 132 mg remain from the original sample.
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?
To solve this problem, we can use the formula for exponential decay:
N = N₀ * (1/2)^(t/t₁/₂)
where N is the final amount, N₀ is the initial amount, t is the time elapsed, and t₁/₂ is the half-life.
a) We are given that after 15.0 minutes, 132 mg remain from the original sample. Therefore, N = 132 mg, and t = 15.0 minutes. We are asked to find the original mass, which is N₀.
132 = N₀ * (1/2)^(15.0/8.51)
Rearranging the equation, we get:
N₀ = 132 * (2)^(15.0/8.51)
Calculating this expression gives us:
N₀ ≈ 132 * 1.901
N₀ ≈ 251.35 mg
Therefore, the original mass of the sample of iron-53 was approximately 251.35 mg.
b) We are given that only 58 mg remain of the original sample. Therefore, N = 58 mg. We are asked to find the time elapsed, which is t.
58 = 251.35 * (1/2)^(t/8.51)
Rearranging the equation, we get:
t/8.51 = log(58/251.35) / log(1/2)
t ≈ 8.51 * (log(58/251.35) / log(1/2))
Calculating this expression gives us:
t ≈ 8.51 * (-0.858) ≈ -7.35 minutes
Since time cannot be negative, there must be an error in the calculation. Please recheck the given information or the calculation.