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

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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:

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

$sf(-2.7724=-0.08663t)$

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