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Chapter 21: Problem 69

A laboratory rat is exposed to an alpha-radiation source whose activity is \(14.3 \mathrm{mCi}\). (a) What is the activity of the radiation in disintegrations per second? In becquerels? (b) The rat has a mass of $385 \mathrm{~g}\( and is exposed to the radiation for \)14.0 \mathrm{~s}$, absorbing \(35 \%\) of the emitted alpha particles, each having an energy of $9.12 \times 10^{-13} \mathrm{~J}$. Calculate the absorbed dose in millirads and grays. (c) If the RBE of the radiation is \(9.5,\) calculate the effective absorbed dose in mrem and Sv.

Short Answer

Expert verified
The activity of the alpha radiation source is \(5.32\times 10^7\) disintegrations/s (Bq). The absorbed dose by the rat is 408 mrad or 4.08 Gy. The effective absorbed dose is 3876 mrem or 38.76 Sv.

Step by step solution

01

Part (a): Activity Conversion

First, we need to convert the activity of the alpha radiation source from mCi to disintegrations per second and becquerels. The activity in mCi is given as \(14.3\mathrm{~mCi}\). 1 Ci is equal to \(3.7\times10^{10}\) disintegrations per second: 1. Convert mCi to Ci: \(14.3\mathrm{~mCi} = 14.3 \times 10^{-3} \mathrm{~Ci}\) 2. Convert Ci to disintegrations per second: \(14.3 \times 10^{-3} \mathrm{~Ci} \times 3.7\times10^{10}\mathrm{~disintegrations/s.Ci} = N \mathrm{~disintegrations/s}\) 3. Convert disintegrations per second to becquerels (1 Bq = 1 disintegration/s): \(N\mathrm{~disintegrations/s} = N\mathrm{~Bq}\)
02

Part (b): Absorbed Dose Calculation

1. Calculate the number of alpha particles absorbed by the rat: - Total alpha particles emitted: \(N \times 14.0 \mathrm{~s}\) - Absorbed alpha particles: \(35\% \times \mathrm{total~alpha~particles~emitted}\) 2. Calculate the energy absorbed by the rat: - Absorbed energy = Absorbed alpha particles × Energy per alpha particle 3. Calculate the absorbed dose in millirads: - Absorbed dose (rad) = \(\frac{\mathrm{Absorbed~Energy}}{\mathrm{Mass~of~the~rat}}\) - Absorbed dose (mrad) = Absorbed dose (rad) × \(10^3\) 4. Convert millirads to grays: - Absorbed dose (Gy) = \(\frac{\mathrm{Absorbed~dose~(mrad)}}{100}\)
03

Part (c): Effective Absorbed Dose Calculation

1. Calculate the effective absorbed dose in mrem: - Effective absorbed dose (rem) = RBE × Absorbed dose (rad) - Effective absorbed dose (mrem) = Effective absorbed dose (rem) × \(10^3\) 2. Convert mrem to Sv: - Effective absorbed dose (Sv) = \(\frac{\mathrm{Effective~absorbed~dose~(mrem)}}{100}\)

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Most popular questions from this chapter

Each statement that follows refers to a comparison between two radioisotopes, \(A\) and \(X .\) Indicate whether each of the following statements is true or false. (a) If the half-life for \(\mathrm{A}\) is shorter than the half-life for \(\mathrm{X}, \mathrm{A}\) has a larger decay rate constant. (b) If \(X\) is "not radioactive," its half-life is essentially zero. (c) If A has a half-life of 10 yr, and \(X\) has a half-life of $10,000 \mathrm{yr}$, A would be a more suitable radioisotope to measure processes occurring on the 40 -yr time scale.

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Methyl acetate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right)\) is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labeled with oxygen-18, the oxygen-18 ends up in the methyl acetate: (a) Do the \(\mathrm{C}-\mathrm{OH}\) bond of the acid and the \(\mathrm{O}-\mathrm{H}\) bond of the alcohol break in the reaction, or do the \(\mathrm{O}-\mathrm{H}\) bond of the acid and the \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol break? (b) Imagine a similar experiment using the radioisotope \({ }^{3} \mathrm{H}\), which is called tritium and is usually denoted T. Would the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) provide the same information about which bond is broken as does the above experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\) ?

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