Many radioisotopes have important industrial, medical, and research applications
ID: 2021225 • Letter: M
Question
Many radioisotopes have important industrial, medical, and research applications. One of these is 60Co, which has a half-life of 5.20 years and decays by the emission of a beta particle (energy 0.310 MeV) and two gamma photons (energies 1.17 MeV and 1.33 MeV). A scientist wishes to prepare a 60Co sealed source that will have an activity of at least 11.9 Ci after 44.0 months of use.If the activity is 11.9 Ci, how many 60Co atoms are in the source?
Tries 0/10
What is the minimum number of nuclei in the source at the time of creation?
Tries 0/10
What is the minimum initial mass of 60Co required?
Explanation / Answer
half life of 60Co t1/2 = 5.2 years = (5.2*31536000) s = 163987200 s final activity A = 11.9 Ci = 11.9*3.7*1010 Bq = 4.403*1011 Bq time t = 30 months = 7.776*107 s decay constant = 0.693 / t1/2 = 0.693 / (163987200 s) = 4.226*10-9 s-1 using Rutherford Soddy law , A = A0 e-t A0 = A et = (4.403*1011 Bq)e( 4.226*10^-9*7.776*10^7) = 6.11*1011 = 6.11*1011 ..................................................................... number of nuclei N0 = A0/ = (6.11*1011) / (4.226*10-9 s-1) = 1.44*1020 ..................................................................... mass of substance m = {(60 gm)(N0)}/ NA = (60*10-3 kg)(1.44*1020 ) / (6.023*1023 ) = 14.4*10-6 kg = 14.4*10-3 gm = (60*10-3 kg)(1.44*1020 ) / (6.023*1023 ) = 14.4*10-6 kg = 14.4*10-3 gmRelated Questions
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