Question

Begin with 800 mCi of 117I. About how much would b left in about 2 hrs?

Answer 1

for 117l t1/2 = 133sec
it is first order
rate constant K = 0.693/t1/2
                = 0.693/133 = 0.00521sce^-1
K = 2.303/tlog(a/a-x)
t= 2hr = 2*60*60 =7200 sec
0.00521 = 2.303/7200log(a/800)
log(a/800) = -0.00521*7200/2.303
log(a/800) = -16.288
    a/800 = 10^-16.288
          = 5.15*10^-17
          = 4.12*10^-14 mCi