A radioactive substance decays at a rate of y = ae^-0.1483t,

Question

A radioactive substance decays at a rate of y = ae^-0.1483t, where t in in hours. Find the half-life of the substance.

 

a. t = 0.8622 years

b. t = 4.674 years

c. t = 18.323 years

d. t = 0.1028 years

 

Answer 1

y = ae^-0.1483t

 

When t = 0, y = a = original amount

 

To find the half-life, we must find t so that y = (1/2)a

 

(1/2)a = ae^-0.1483t

 

(1/2) = e^-0.1483t

 

Take natural log of both sides:  ln(0.5) = ln(e^-0.1483t)

 

                                                       ln(0.5) = -0.1483t

 

                                                                t = ln(0.5)/(-0.1483)

 

                                                                  = 4.674 years