In 3 years, 30% of a radioactive element decays. Find its half-life.

Question

In 3  years, 30%  of a radioactive element decays. Find its half-life.

Answer 1

A(t) = A₀e^kt = amount remaining after t years

 

 (A₀ = initial amount and k is a constant to be determined)

 

We are given: A(3) = 0.70A₀ (if 30% decays, then 70%                                                                       remains),               

      0.70A₀ = A₀e^3k

 

      0.70 = e^3k

 

      ln(0.70) = 3k    So, k = -0.11889

 

A(t) = A₀e^-0.11889t

 

To determine the half-life, we must find t so that A(t) = 0.50A₀

 

So, we have 0.50A₀ = A₀e^-0.11889t

 

                    0.50 = e^-0.11889t

 

                    ln(0.50) = -0.11889t

 

                             t = 5.83 years 

 

 

Answer 2

A = a₀(0.5)^t/h, where a₀ = starting amount, t = time in years, h = half-life in years.

 

0.30a₀ = a₀(0.5)^3/h                           divide both sides by a₀

0.3 = (0.5)^3/h                                   take natural log of both sides

ln 0.3 = (3/h) ln 0.5                         calculate

-1.2039728 ≈ (3/h)(-0.693147181  multiply both sides by h

-1.2039728h ≈ 3(-0.693147181)    calculate

-1.2039728h ≈ -2.07944154           divide both sides by left coefficient

h ≈ 1.72714993

 

The half-life is approximately 1.7 years.