Question #249015

1.    A radioactive sample with a half-life of  3hrs contains 25-mCi radioactivity. The radioactivity of the sample after 9 days days would be?

2.1.    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.

 


1
Expert's answer
2021-10-10T16:02:12-0400

1. 9 days = 9×\times24 = 216 hours, which 2163=\frac{216}{3} = 72 half-life periods

The activity will be 25272=5.29×1021mCi\frac{25}{2^{72}} = 5.29 \times10^{-21} mCi

2.

a) 1 microgram is 10610^{-6} g; the number of atoms is 106×6.022×1023131=4.60×1015\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 125,000106=0.125\frac{125,000}{10^6}=0.125 curie

c) 50 days is 508.02=6.23\frac{50}{8.02}=6.23 half-life periods

The activity will be 0.12526.23=0.00167\frac{0.125}{2^{6.23}}=0.00167 curie

d)

logaa0=ln2t1/2t=106tt=logaa0106log0.00010.125=3.097t=3.097106=3.097×106 slog\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 3.097×24×36003.097\times24\times3600 = 35.8 days


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