Question 1129784

{{{A = A[0]*e^(-kt) }}}


given: 

{{{A[0]=0.3}}}

Since the decay rate is {{{1.15}}}% per day, the amount of I-125 left after {{{t = 1 }}}day is:  

 {{{A = 0.3 - (1.15/100)*0.3 g = 0.3-0.045= 0.29655g}}}


substituting for A, A₀ and t in eq(1) we get:

{{{0.29655 = 0.3*e^(-k*1) }}}

{{{ln(0.29655) =ln( 0.3*e^(-k*1) )}}}

{{{ln(0.29655) =ln( 0.3)+ln(e^(-k*1) )}}}

{{{ln(0.29655) -ln( 0.3)=ln(e^(-k*1) )}}}

{{{ln(e^(-k) )=ln(0.29655/0.3) 

{{{-k*ln(e )=ln(0.9885)}}} .........{{{ln(e)=1}}}

{{{-k=-0.0115666}}}

{{{k=0.0115666}}}


To the nearest day, how long will it take for half of the Iodine-125 to decay? 


{{{0.29655/ 2= 0.3*e^(-0.0115666*t)}}}

{{{t=60.9266}}}
 
{{{t}}}≈ {{{61}}} days