Half-life question

Question

It is observed that the activity of the radioactive element decreases exponentially according to the formula A(t) = A(10)^-kt, and decreases to 0.35 of its original quantity in 5h. Find the half-life of the element.

Answer 1

A(t) = A(10)^-kt

0.35 = 10^-kt

log 0.35 = log 10^-kt

log 0.35 = -kt log 10

log 0.35 = -kt

log 0.35 = -5k

-0.455931 ≈ -5k

-0.091186 ≈ -k

0.091186 ≈ k

 

0.5 ≈ 10^-0.091186t

log 0.5 ≈ log 10^-0.91186t

log 0.5 ≈ -091196t

-0.30102 ≈ -0.91196t

3.3008 ≈ t

 

The half life is approximately 3.3 hours.