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 16.5 Ci after 25.0 months of use. If the activity is 16.5 Ci, how many 60Co atoms are in the source? 1.45×1020 You are correct. What is the minimum number of nuclei in the source at the time of creation? 1.91×1020 You are correct. What is the minimum initial mass of 60Co required?

Explanation / Answer

1Ci = 3.7×10^10 s¹
So the desired activity is A = 16.5 Ci = 6.105×10¹¹ s¹ (decays per second)

The activity is is the number of atoms decaying per second: A = -dN/dt
Radioactive decay follows the rate law:   -dN/dt= ·N   ( decay constant)

Solving this differential equation you find Number of atoms as function of time:
N = N·e^(-·t) (Nnumber of molecules at t=0)
From the equation above you can easily show that half life and decay constant ar related as:
= ln(2)/t
So the number of atoms in terms of time and half life is given by: N = N·e^(-(ln(2)/t)·t) = N·(1/2)^(t/t)

Combine the expression and solve for initial number of atoms: A = ·N
<=> A = [ln(2)/t] · N · (1/2)^(t/t)
<=> N = A· 2^(t/t) · t / ln(2)
Then multiply by atomic mass to obtain initial mass m = M·N = M · A · 2^(t/t) with M = 59.934u
=> m= 59.933817 *1.66055×10²kg * 6.105 ×10¹¹ s¹ * 2^( 25/ 5.2*12month) * (5.2*365*24*3600s) / ln(2)
= 1.898 ×10 kg
= 18.98 mg

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