   Chapter 32, Problem 21PE

Chapter
Section
Textbook Problem

Large amounts of 65Zn are produced in copper exposed to accelerator beams. While machining contaminated copper, a physicist ingests 50.0 μCi of 65Zn. Each 65Zn decay emits an average γ −ray energy of 0.550 MeV, 40.0% of which is absorbed in the scientist’s 75.0−kg body. What dose in mSv is caused by this in one day?

To determine

The radiation dose absorbed in the scientist's body in one day after he ingests 50.0µCi of 65Zn.

Explanation

Given:

The activity of 65Zn

R=50.0 μCi

Average energy of a ϒ ray

Eγ=0.550 MeV

Percentage of energy absorbed

40.0% of the total energy

Mass of the scientist's body

m=75.0 kg

Calculation:

Express the activity of 65Zn in Becquerel (Bq)

R=50.0 μCi× 10 6Ci1μCi×3.7× 10 10Bq1 Ci=1.850×106Bq=1.850×106 disintegrations/s

Each disintegration releases 1 ϒ ray of average energy Eγ . Calculate the total energy released per second.

E/s=REγ=1.850× 106 disintegrations/s0.550 MeV=1.0175×106MeV/s

Express the energy released per second in J/s.

E/s=1.0175× 106MeV/s 1.6× 10 13 J 1 MeV=1.628×107J/s

Calculate the energy released in a day.

E=1.628× 10 7J/s 3600 s 1 h 24 h 1 d=1.40659×102J/day

The energy absorbed by the body of the scientist is given by,

Eabs=40

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