Answer to Question #249294 in Atomic and Nuclear Physics for Yts

Question

A sample of material contains 1 microgram of iodine-131. Note that, iodine-131 plays a major role as a radioactive isotope present in nuclear and it a major contributor to the health hazards when released into the atmosphere during an accident. Iodine-131 has a half-life of 8.02 days.

a)The number of iodine-131 atoms initially present.

b)The activity of the iodine-131 in curies.

c)The number of iodine-131 atoms that will remain in 50 days.

d)The time it will take for the activity to reach 0.1 mCi.

Answer 1

a) 1 microgram is $10^{-6}$ g; the number of atoms is $\frac{10^{-6}\times6.022\times10^{23}}{131}=4.60\times10^{15}$

b) The activity of iodine-131 per gram is known to be 125,000 curie; thus, the activity of 1 mg is $\frac{125,000}{10^6}=0.125$ curie

c) 50 days is $\frac{50}{8.02}=6.23$ half-life periods

The activity will be $\frac{0.125}{2^{6.23}}=0.00167$ curie

d)

$log\frac{a}{a_0}=-\frac{ln2}{t_{1/2}}t=-10^{-6}t \\ t=-\frac{log\frac{a}{a_0}}{10^{-6}} \\ log\frac{0.0001}{0.125}=-3.097 \\ t=\frac{3.097}{10^-6}=3.097\times10^6 \ s$ or $3.097\times24\times3600$ = 35.8 days