Half Life word problem

Question

A 64g sample of Germanium-66 is left undisturbed for 12.5 hours. At the end of the period, only 2.0g remain. What is the half- life of this material. Please use the formula A= A₀ (1/2) ^(t/H)

Answer 1

A(t) = A₀(1/2)^t/H

 

A(t) = amount remaining after time t hours

H = half-life in hours

A₀ = initial amount present at time t = 0 hours

 

Take natural of both sides and solve for H:

 

ln(A/A₀) = (t/H) ln(1/2)

 

H = t ln(1/2) / ln(A/A₀)

 

H = (12.5 hours) ln(1/2) / ln(2.0/64) = 2.5 hours

Answer 2

Well, you started with 64g and ended with 2g, which is 1/(2⁵) of the original. That is, we multiplied the mass by 1/2 five times in 12.5 hours. So the half-life must have elapsed 5 times in 12.5 hours. So what does that give for the half-life?