If 20% of the substance disappears in 10 years, what is its half-life? (Round your answer to two

Question

A radioactive substance follows the decay equation dA/ dt= kA.
If 20% of the substance disappears in 10 years, what is its half-life? (Round your answer to two decimal places.)

Answer 1

First step is to integrate:

 

dA/A = k dt

 

ln(A) = kt + C₁

 

A(t) = C₀e^kt

 

At t=0, we have the initial amount of the substance. Call this amount A₀.

 

A(0) = A₀ = C₀e⁰ = C₀, so we know that our constant of integration is simply the initial amount: A(t) = A₀e^kt.

 

Next we use the second bit of information given in the problem. At t=10 years, we've lost 20% of the starting amount, which leaves 0.8A₀ behind:

 

A(10) = 0.8A₀ = A₀e^10k. Now we can solve for k.

 

10k = ln(0.8)

k= (1/10)*ln(0.8).

 

Finally, we can solve for half life T. At t=T, half the starting amount will remain, A(T)=0.5A₀:

 

0.5 A₀ = A₀e^kT

kT = ln(0.5)

T = ln(0.5)/k = 10 ln(0.5)/ln(0.8)