Question

In the manufacture of steel engine components radioactive iron (59Fe) is included in a total mass of 0.20 Kg. The component is placed in a test engine with surrounding oil lubricant when the activity due to the isotope is 20.0

Answer 1

half life of 59Fe       t_1/2= 45.1 days

                               t_1/2= 45.1 x 24     = 1082.4 hours

disintegration constant    lamda= 0.693/t_1/2

                                                    =0.693/1082.4         =6.4x10^-4 hour^-1

the activity of 59Fe   R₀=20x10^-6ci                                     1ci=3.7x10¹⁰ disintegration per second

                              R₀= 20x10^-6 x3.7x10¹⁰/(60 x60)

                                       =205.56 disintegration per hour

Activity of 0.5 litre oil contain 59Fe after 1000hr     R= R₀ e^{-lamda x t}

                                       R= 205.56 x e^{-6.4x10-4x 1000}

                                                       R= 108.39 disintegration per hr

the activity of lubricant oil removed from engine contain 59Fe to produce 800disintegration per minute per litre after 1000 hour

activity after 1000 hour

                           r= 800/1000

                               = 0.8 disintegration per hour per litre

                                  =0.8/2 =0.4 disintegration per hour per 0.5 litre

Mass of steel component M=0.20kg

the mass of 59Fe worn out from engine

        m= r x M / R

                        =0.4 x 0.2/ 205.56

                         =3.89x10^-4

                          the mass of 59Fe worn out from steel component is 3.89x10^-4kg or 0.389mg