Many radioisotopes have important industrial, medical, and research applications

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 15.5 Ci after 45.0 months of use.
If the activity is 15.5 Ci, how many 60Co atoms are in the source?

What is the minimum number of nuclei in the source at the time of creation?

What is the minimum initial mass of 60Co required?

Explanation / Answer

60Co
half life = 5.2 years = t1/2
energy of beta prticel, Eb = 0.310 MeV
energy of photons, E1 = 1.17 MeV , E2 = 1.33 MeV

now, N = Noe^(-lambda*t)
Activity A = dN/dt = -N*lambda
A = 15.5 Ci = 5.735*10^11 Bq = N*ln(2)/5.2*365*24*60*60
N = 1.3568*10^20 atoms

at the tie of creation , numebr of nuclei = No
N = No*e^(-lambda*45/12)
lamdba = ln(2)/5.2
No = 2.23668*10^20 atoms

minimum initial mass = No*59.9338/6.022*10^23 grams = 0.02226 grams

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