Question

Gamma rays may be used to kill pathogens in ground beef. One irradiation facility uses a 60Co source that has an activity of 1.0×106Ci. 60Co undergoes beta decay and then gives off two gamma rays, at 1.17 and 1.33 MeV; typically 30% of this gamma-ray energy is absorbed by the meat. The dose required to kill all pathogens present in the beef is 4000 Gy.
How many kilograms of meat per hour can be processed in this facility? Express your answer in kilograms per hour.

Answer 1

Answer:

Explanation:

$C_i=3.7\times 10^{16} \,decays/sec$

Energy of gamma rays due to ore decay is

$(1.17+1.33)MeV=2.50MeV$

Energy of gamma ray produced in seconds is

$(2.5\times 3.7\times 10^{16})MeV$

Activity

$1 \times 10^6c$

Total energy of gamma ray produced in second is

$(2.5 \times 3.7 \times 10^{16} \times 10^6)MeV$

E-energy of gamma ray produced in an hour is

$(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV$

Gamma ray energy absorbed by meat = 30% (in 1 hour) of E is 0.3E

Dose required to kill pathogen = 4000Gy=4000J/kg

The kilogram of meat that can be produced per hour is

$\frac{0.3E/hr}{4000J/kg}\\\\E=(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV\\\\=333\times 10^{18}MeV\\\\eV=1.6\times 10^{-19}\\\\E=53.3J$

The meat that can be produced =$\frac{0.3\times 53.3}{4000}=3.996\times 10^{-3}$