Question

A 7.85 × 10-5 mol sample of copper-61 emits 1.47 × 1019 positrons in 90.0 minutes. what is the decay constant for copper-61

Answer 1

4.14x10^-3 per minute
First, calculate how many atoms of Cu-61 we initially started with by
multiplying the number of moles by Avogadro's number.

7.85x10^-5 * 6.0221409x10^23 = 4.7273806065x10^19
Now calculate how many atoms are left after 90.0 minutes by subtracting the
number of decays (as indicated by the positron emission) from the original
count.
4.7273806065x10^19 - 1.47x10^19 = 3.2573806065x10^19
Determine the percentage of Cu-61 left.
3.2573806065x10^19/4.7273806065x10^19 = 0.6890455577
The formula for decay is:
N = N0 e^(-λt)
where
N = amount left after time t
N0 = amount starting with at time 0
λ = decay constant
t = time
Solving for λ:
N = N0 e^(-λt)
N/N0 = e^(-λt)
ln(N/N0) = -λt
-ln(N/N0)/t = λ
Now substitute the known values and solve:
-ln(N/N0)/t = λ
-ln(0.6890455577)/90m = λ
0.372447889/90m = λ
0.372447889/90m = λ
0.00413830987 1/m = λ
Rounding to 3 significant figures gives 4.14x10^-3 per minute as the decay
constant.