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 13.2 Ci after 45.0 months of use.
a) If the activity is 13.2 Ci, how many 60Co atoms are in the source?

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

c) What is the minimum initial mass of 60Co required?

Explanation / Answer

From the given data

The half life of 60Co t1/2 = 5.2 years = (5.2*31536000) s

                                                 = 163987200 s

final activity A = 13.2 Ci

                      = 13.2*3.7*1010 Bq

                      = 4.884*1011 Bq

time t = 45 months = 1.1664*108 s

then decay constant = 0.693 / t1/2

                          = 0.693 / (163987200 s)

                          = 4.226*10-9 s-1

Now using Rutherford Soddy law ,

                  A = A0 e-t

                 A0 = A et

                       = (4.884*1011 Bq)e( 4.226*10^-9*1.664*10^8)

                       = 9.864*1011

Therefore number of nuclei N0 = A0/

                                = (9.864*1011) / (4.226*10-9 s-1)

                                = 2.334*1020

mass of substance m = {(60 gm)(N0)}/ NA

                                = (60*10-3 kg)(2.334*1020 ) / (6.023*1023 )

                               = 23.25*10-6 kg

                               = 23*10-3 grams

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.