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1 month ago

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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.

1 answer:

ValentinkaMS [3.4K]1 month ago

4 0

Answer:

Explanation:

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

The energy from gamma rays generated by ore decay is

$(1.17+1.33)MeV=2.50MeV$

The energy from gamma rays created per second is

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

Activity

$1 \times 10^6c$

The total energy of gamma rays produced in one second is

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

E represents the energy of gamma rays generated in one 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

The dosage necessary to eliminate pathogens = 4000Gy=4000J/kg

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

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According to the problem, we know that

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