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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
(a) To convert the activity to disintegrations per second and becquerels: $$14.3\ \mathrm{mCi} \times \frac{1\ \mathrm{Ci}}{1000\ \mathrm{mCi}} \times 3.7 \times 10^{10}\ \mathrm{Bq} = 5.3 \times 10^{8}\ \mathrm{Bq}$$ (b) Absorbed energy: $$E_{absorbed} = 5.3 \times 10^{8}\ \mathrm{disintegrations/s} \times 9.12 \times 10^{-13}\ \mathrm{J/disintegration} \times 14 \ \mathrm{s} \times 0.35 = 2.368 \times 10^{-3}\ \mathrm{J}$$ Absorbed dose: $$D = \frac{2.368 \times 10^{-3}\ \mathrm{J}}{385\ \mathrm{g}} = 6.144 \times 10^{-3}\ \mathrm{J/g}$$ Convert to millirads and grays: $$6.144 \times 10^{-3}\ \mathrm{J/g} \times 100 \mathrm{mrad/J} = 614.4\ \mathrm{mrad}$$ $$6.144 \times 10^{-3}\ \mathrm{J/g} \times 1000\ \mathrm{g/kg} \times 0.01\ \mathrm{Gy/J} = 0.06144\ \mathrm{Gy}$$ (c) Effective absorbed dose: $$D_{eff} = 614.4\ \mathrm{mrad} \times 9.5 = 5,837.6\ \mathrm{mrem}$$ $$D_{eff} = 0.06144\ \mathrm{Gy} \times 9.5 = 0.58368\ \mathrm{Sv}$$

Step by step solution

01

(a) Convert the activity to disintegrations per second and becquerels

To convert the activity from \(\mathrm{mCi}\) to disintegrations per second and becquerels, use the conversion factors \(1\mathrm{Ci} = 3.7 \times 10^{10}\ \mathrm{Bq}\) and \(1\mathrm{Bq} = 1\ \mathrm{disintegration/second}\). Given, the activity is \(14.3\ \mathrm{mCi}\). To convert to disintegrations per second and becquerels: $$14.3\ \mathrm{mCi} \times \frac{1\ \mathrm{Ci}}{1000\ \mathrm{mCi}} \times 3.7 \times 10^{10}\ \mathrm{Bq}$$
02

(b) Calculate the absorbed energy

Firstly, we need to calculate the energy absorbed. The absorbed energy can be found using the formula: $$E_{absorbed} = N \times E_{\alpha} \times T \times \%$$ Here, \(N\) represents the activity of the radiation in disintegrations per second, \(E_{\alpha}\) is the energy per alpha particle, \(T\) is the exposure time in seconds, and \(\%\) is the absorbed percentage of alpha particles.
03

(b) Calculate the absorbed dose in millirads and grays

To find the absorbed dose, divide the absorbed energy by the mass of the rat and use the conversion factors \(1 \mathrm{rad} = 0.01\ \mathrm{Gy}\) and \(1\ \mathrm{Gy} = 1\ \mathrm{J/kg}\). The formula is: $$D = \frac{E_{absorbed}}{m}$$ Here, \(D\) represents the absorbed dose, \(E_{absorbed}\) is the absorbed energy we calculated above, and \(m\) is the mass of the rat.
04

(c) Calculate the effective absorbed dose in mrem and Sv

To find the effective absorbed dose, multiply the absorbed dose by the RBE of the radiation. Then, convert the dose from rads to mrem and from grays to Sv. The formula for effective absorbed dose is: $$D_{eff} = D \times RBE$$ Here, \(D_{eff}\) represents the effective absorbed dose, \(D\) is the absorbed dose we calculated earlier, and \(RBE\) is the Relative Biological Effectiveness. Using the conversion factors \(1 \mathrm{mrad} = 0.001\ \mathrm{rad}\) and \(1 \mathrm{mSv} = 0.001\ \mathrm{Sv}\), we can convert the effective absorbed dose to mrem and Sv.

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Most popular questions from this chapter

Based on the following atomic mass values \(-{ }^{1} \mathrm{H}\), \(1.00782 \mathrm{amu} ;{ }^{2} \mathrm{H}, 2.01410 \mathrm{amu} ;{ }^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;{ }^{3} \mathrm{He}\), \(3.01603\) amu; \({ }^{4} \mathrm{He}, 4.00260\) amu-and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1} \mathrm{H}\)

Tests on human subjects in Boston in 1965 and 1966 , following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits \(8 \times 10^{-13} \mathrm{~J}\) of energy and if the average person weighs \(75 \mathrm{~kg}\), calculate the number of rads and rems of radiation in \(1 \mathrm{yr}\) from such a level of plutonium.

A sample of an alpha emitter having an activity of \(0.18 \mathrm{Ci}\) is stored in a 25.0-mL sealed container at \(22^{\circ} \mathrm{C}\) for 245 days. (a) How many alpha particles are formed during this time? (b) Assuming that each alpha particle is converted to a helium atom, what is the partial pressure of helium gas in the container after this 245 -day period?

Write balanced equations for each of the following nuclear reactions: (a) \({ }_{92}^{238} \mathrm{U}(\mathrm{n}, \gamma)^{239} \mathrm{G}_{2} \mathrm{U}\), (b) \({ }_{7}^{14} \mathrm{~N}(\mathrm{p}, \alpha){ }_{6}^{11} \mathrm{C}\), (c) \({ }^{18} \mathrm{O}(\mathrm{n}, \beta)_{9}^{19} \mathrm{~F}\).

The average energy released in the fission of a single uranium-235 nucleus is about \(3 \times 10^{-11} \mathrm{~J}\). If the conversion of this energy to electricity in a nuclear power plant is \(40 \%\) efficient, what mass of uranium- 235 undergoes fission in a year in a plant that produces \(1000 \mathrm{MW}\) (megawatts)? Recall that a watt is \(1 \mathrm{~J} / \mathrm{s}\).

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