# 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?

May 21, 2017

32 days.

#### Explanation:

The expression for 1st order decay is:

sf(N_t=N_0e^(-lambdat)

$\textsf{\lambda = \frac{0.693}{t} _ \left(\frac{1}{2}\right) = \frac{0.693}{8.0} = 0.08663 \textcolor{w h i t e}{x} {d}^{- 1}}$

Taking natural logs of both sides:

$\textsf{\ln {N}_{t} = \ln {N}_{0} - \lambda t}$

$\therefore$$\textsf{\ln \left({N}_{t} / {N}_{0}\right) = - \lambda t}$

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

$\textsf{\ln \left(\frac{6.22 \times {10}^{17}}{9.95 \times {10}^{18}}\right) = - 0.08663 t}$

$\textsf{- 2.7724 = - 0.08663 t}$

$\textsf{t = \frac{2.7724}{0.08663} = 32 \textcolor{w h i t e}{x} \text{days}}$