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 58g remain of the original sample, how much time has elapsed?
a) To find the original mass of the sample, we can use the formula for exponential decay:
N(t) = N0 * (1/2)^(t/T)
where N(t) is the amount remaining after time t, N0 is the initial amount, and T is the half-life.
Let's plug in the given values:
N(t) = 132 mg
N0 = ?
T = 8.51 minutes
t = 15.0 minutes
132 mg = N0 * (1/2)^(15.0/8.51)
To isolate N0, we can divide both sides of the equation by (1/2)^(15.0/8.51):
N0 = 132 mg / (1/2)^(15.0/8.51)
Using a calculator, we can determine that (1/2)^(15.0/8.51) is approximately 0.145.
Therefore, the original mass of the sample of iron-53 is:
N0 = 132 mg / 0.145 ≈ 910.34 mg
Hence, the original mass of the sample of iron-53 was approximately 910.34 mg.
b) To find how much time has elapsed when only 58 g remain of the original sample, we can rearrange the formula for exponential decay and solve for t:
N(t) = N0 * (1/2)^(t/T)
Let's plug in the given values:
N(t) = 58 g
N0 = 910.34 mg (converted to grams = 0.91034 g)
T = 8.51 minutes
t = ?
58 g = 0.91034 g * (1/2)^(t/8.51)
To isolate t, we can divide both sides of the equation by 0.91034 g:
(58 g) / (0.91034 g) = (1/2)^(t/8.51)
Using a calculator, we can determine that (58 g) / (0.91034 g) is approximately 63.655.
Therefore, the time that has elapsed when only 58 g remain of the original sample is:
(1/2)^(t/8.51) = 63.655
We can solve this equation by taking the logarithm of both sides. Taking the logarithm base 2 (since we have (1/2) on the right side):
(t/8.51) * log2(1/2) = log2(63.655)
The logarithm of 1/2 base 2 is -1, and the logarithm of 63.655 base 2 is approximately 5.997.
(t/8.51) * (-1) = 5.997
To isolate t, we can multiply both sides of the equation by -8.51:
t = -8.51 * 5.997 ≈ -51.02
Since time cannot be negative, we can discard this result.
Therefore, there is no valid solution for the elapsed time when only 58 g remain of the original sample.