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 
 Î» = 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.