Consider a 5.28 gram sample of sodium iodide (NaI), of which .0160% of the iodin
ID: 2007333 • Letter: C
Question
Consider a 5.28 gram sample of sodium iodide (NaI), of which .0160% of the iodine having a half life of 13 days. Take the atomic weights of Na and the remainder of the I to be 23.0 and the 126.9 amu, Respectively.a. Determine the number of radioactive I-126 atoms in the sample.
b. What is the initial activity of the sample?
c. What is the activity of the sample after 147.0 days?
d. When will the activity of the sample be 6.88 micro ci?
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Explanation / Answer
GIven mass of sample 5.28 grams half life of the sample is 13 days convert mass of sample from grams to moles number of moles n = 5.28 / 126.9 = 0.04160 moles Number of I - 126 atoms in the sample is N = n A v = 0.04160 * 6.02 *10 23 = 2.5*1022 atoms number os sodium iodide atoms in the sample No = ( 5.28 / ( 23+126.9 ) ) ( 6.02 *10 23 ) = 2.120*1022 atoms decay constant of Na I is = 0.693 / t 1/2 = 0.693 / 13 = 0.05330 dayss -1 b) initial activity of the sample A = No = ( 2.120* 10 22 ) ( 0.05330) = 0.1130 * 10 22 decays / sec = 3 *10 10 C i c) number of nuclei after 147 days N = N o ( e -t ) = ( 2.120*1022 ) ( e - ( 0.05330)(147) ) = 8.3867*1018 nuclei Activity of the sample after 147 days A = N = ( 8.3867*1018 ) ( 0.05330) = 4.47*1017 dacys / sec = 1.2*107 C i d) Activity of the sample is 6 Ci A = N N = 6 * 10 -6 * 3.7*1010 decays / sec / (0.05330 s-1 ) = 4.165*106 atoms N = N o e -t e t = N o / N = 2.120*1022 / 4.165*106 = 5*1015 t = ln ( 5*1015 ) t = ln ( 5*1015 ) / = 36.1482/0.05330 = 678.2 daysRelated Questions
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