Iodine-131, #t_(1/2)# = 8.0 days, is used in treatment of thyroid gland diseases. If a sample of iodine-131 initially emits #9.95 x 10^18 beta# particles per day, how long will it take for the activity to drop to #6.22 x 10^17 beta# particles per day?

Question

5046 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-21T16:18:30

32 days.

The expression for 1st order decay is:

#sf(N_t=N_0e^(-lambdat)#

#sf(lambda=0.693/t_(1/2)=0.693/8.0=0.08663color(white)(x)d^(-1))#

Taking natural logs of both sides:

#sf(lnN_t=lnN_0-lambdat)#

#:.##sf(ln(N_t/N_0)=-lambdat)#

The activity of a sample is proportional to the number of undecayed atoms so we can write:

#sf(ln((6.22xx10^(17))/(9.95xx10^(18)))=-0.08663t)#

#sf(-2.7724=-0.08663t)#

#sf(t=2.7724/0.08663=32color(white)(x)"days")#