Answer to Question #100903 in Atomic and Nuclear Physics for saif alwash

Question

The radioactive nuclei 60Co is widely used in medical applications. It undergoes
beta decay, and the total energy of the decay process is 2.82 MeV per decay event. The
half-life of this nucleus is 272 days. Suppose that a patient is given a dose of 6.9 µCi of
60Co. If all of this material decayed while in the patient's body, what would be the total
energy deposited there?

Answer 1

$1Ci=$ $3.7*10^{10} deacys/sec$

Number of decays per second in 6.9$\mu$ Ci of cobalt-60$=6.9*10^{-6}*3.7*10^{10}=25.53*10^4$

As, it takes approximately 5 half life for the complete decaying of a radioactive nucleus

So total number of decays in time span of 5 half life $=5*272*24*60*60*25.53*10^4=3*10^{13}$

energy released per decay event is 2.72 MeV or 2.72*10⁶ eV

so, energy released in time span of 5 half life =3*10¹³ *2.72*10⁶ eV= 8.16*10¹⁹ eV

The total energy deposited is 8.16*10¹⁹ eV