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 13.7 Ci after 37.0 months of use. If the activity is 13.7 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?

Answer 1

Part 1, minimum number of nuclei in the source at the time of creation

R = λ* N ------------------(1) ; λ - decay constant,

Since you are not given the value of L, you have to find it using,

λ = ln(2) / t1/2 ; where t1/2 is Half Life -------------------(2)

(1) and (2) gives,

R = ln(2) * N / t1/2

=>

N = R * t1/2 / ln(2)

Here R = 13.7 Ci and  t1/2 = 5.20 years = 5.2*365*24*60*60 sec

1 Ci = 3.7 *1010 Bq

N = (13.7 * 3.7*1010) * (5.2*365*24*60*60) / (ln(2))

N = 1.20 * 1020

minimum number of nuclei in the source at the time of creation is  N = 1.20 * 1020

.................................................................................................................................................................

Part 2, minimum initial mass of 60Co

Now use the equation,

N = No * exp (-λt)

N is the number we just found, t is the time in seconds (37 months or 37/12 years)

No = N * exp (+λt)

No = N * exp ( ln(2) * t / t1/2 )

No = 1.20 * 1020   * exp (ln (2) * (37/12)/ 5.2)

No = 1.81 * 1020