Many radioisotopes have important industrial, medical, and research applications

Question

Many radioisotopes have important industrial, medical, and research applications. One of these is 60^Co, 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 60^Co sealed source that will have an activity of at least 16.5 Ci after 25.0 months of use. If the activity is 16.5 Ci, how many 60^Co 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 60^Co required?

Explanation / Answer

1)
You need to find the number of atoms when the activity is 16.5Ci, so you have to use the equation,

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

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

L = ln(2) / Half Life -------------------(2)

(1) and (2) gives,

R = ln(2) * N / half_life

=>

N = R * half_life / ln(2)

now substitute values. Here R is given in Ci, so you have to convert it to Bq by multiplying it by 3.7*10^(10), because => 1 Ci = 3.7*10^(10)Bq

N = (16.5 * 3.7*10^(10)) * (5.2*365*24*60*60) / (ln(2))
N = 1.44 * 10^20

2)

Now use the equation,

N = No * exp (-Lt)

N is the number we just found, t is the time in seconds (25 months or 25/12 years)

No = N * exp (+Lt)
No = N * exp ( ln(2) * t / half_life )
No = 1.44 * 10^20 * exp (ln (2) * (25/12)/ 5.2)
No = 1.900 * 10^20

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