Half life problem

Question

If 17% of a radioactive substance decays in 5 hours, what is the half-life of the substance?

Answer 1

The integrated rate law for a first order reaction is:

 

[A] = [A₀] e^-kt   where [A] = concentration of A at time t   and [A₀]= concentration of A at the very beginning (t=0)

 

Here, we know that when t=5hr, [A] = 0.17 [A₀]  (17% of original)

 

Substitute into equation:

 

0.17 [A₀] = [A₀] e^-k(5)

 

Divide both sides by [A₀]

 

0.17 = e^-5k

 

Take natural log of both sides

 

ln 0.17 = -5k

 

solve for k

 

k = - (ln 0.17)/5

 

Repeat for half life substituting original equation with [A] = 0.5 [A₀] and t = T for the half life (half life is the time it takes for the concentration to reduce by half, or 50%)

 

0.5 [A₀] = [A₀] e^-kT

 

Do the same thing (divide both sides by [A₀] and take natural log as before:

 

ln 0.5 = - kT

 

Substitute k from earlier

 

ln 0.5 = - (- (ln 0.17)/5)*T

 

Solve for T

 

-0.693 = -(-(-1.772)/5)*T

0.693 = 0.354 * T

 

T = 1.957 = 2 hours is the half-life