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Chapter 19: Problem 62

Breeder reactors are used to convert the nonfissionable nuclide 238 \(\mathrm{U}\) to a fissionable product. Neutron capture of the 238 \(\mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?

Short Answer

Expert verified
The final fissionable product after 238U undergoes neutron capture and two successive beta decays in a breeder reactor is 239Pu (plutonium). The process can be described as follows: Neutron capture: \(_{92}^{238}\textrm{U} + _0^1n \rightarrow _{92}^{239}\textrm{U}\) First beta decay: \(_{92}^{239}\textrm{U} \rightarrow _{93}^{239}\textrm{Np} + _{-1}^{0}e\) Second beta decay: \(_{93}^{239}\textrm{Np} \rightarrow_{94}^{239}\textrm{Pu} + _{-1}^{0}e\)

Step by step solution

01

Understand neutron capture and beta decay

A breeder reactor works by converting a non-fissionable isotope, in this case, 238U, into a fissionable product. The process starts with neutron capture by 238U, which will increase the atomic mass but retain the same atomic number. After the neutron capture, the isotope undergoes two successive beta decays. In a beta decay, a neutron transforms into a proton, and an electron (beta particle) is emitted. This causes the atomic number to increase by 1 and the atomic mass to remain the same.
02

Neutron capture by 238U

First, we will analyze the neutron capture by 238U. When a neutron is captured by 238U, the atomic number remains the same but the atomic mass increases by 1: Neutron capture: \(_{92}^{238}\textrm{U} + _0^1n \rightarrow _{92}^{239}\textrm{U}\) The resulting isotope after neutron capture is 239U.
03

First beta decay

Now, let's examine the first beta decay. In beta decay, one neutron is converted into a proton and emits an electron. So, the atomic number increases by 1 and the atomic mass remains the same: First beta decay: \(_{92}^{239}\textrm{U} \rightarrow _{93}^{239}\textrm{Np} + _{-1}^{0}e\) The resulting isotope after the first beta decay is 239Np (neptunium).
04

Second beta decay

Finally, we will analyze the second beta decay. Again, a neutron is converted into a proton, with the emission of a beta particle: Second beta decay: \(_{93}^{239}\textrm{Np} \rightarrow_{94}^{239}\textrm{Pu} + _{-1}^{0}e\) The resulting isotope after the second beta decay is 239Pu (plutonium).
05

Identify the final fissionable product

After two successive beta decays following the neutron capture of 238U, the final fissionable product is 239Pu (plutonium).

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Most popular questions from this chapter

There is a trend in the United States toward using coal-fired power plants to generate electricity rather than building new nuclear fission power plants. Is the use of coal-fired power plants without risk? Make a list of the risks to society from the use of each type of power plant.

Photosynthesis in plants can be represented by the following overall equation: $$ 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\mathrm{} Light \mathrm{}}{\longrightarrow} C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ Algae grown in water containing some $^{18} \mathrm{O}\left(\text { in } \mathrm{H}_{2}^{18} \mathrm{O}\right)$ evolve oxygen gas with the same isotopic composition as the oxygen bin the water. When algae growing in water containing only \(^{18} \mathrm{O}\) were furnished carbon dioxide containing \(^{18} \mathrm{O}\) no \(^{18} \mathrm{O}\) was found to be evolved from the oxygen gas produced. What conclusions about photosynthesis can be drawn from these experiments?

A reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is 15.0 seconds. If 199 atoms of bohrium-267 could be synthesized, how much time would elapse before only 11 atoms of bohrium-267 remain? What is the expected electron configuration of elemental bohrium?

Assume a constant \(1^{14} \mathrm{C} /^{12} \mathrm{C}\) ratio of 13.6 counts per minute per gram of living matter. A sample of a petrified tree was found to give 1.2 counts per minute per gram. How old is the tree? (For $^{14} \mathrm{C}, t_{1 / 2}=5730$ years.)

A rock contains 0.688 \(\mathrm{mg}^{206} \mathrm{Pb}\) for every 1.000 \(\mathrm{mg}^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \(^{206}\mathrm{P}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206 \mathrm{P}} \mathrm{b}\) is negligible, calculate the age of the rock. (For $38 \mathrm{U}, t_{1 / 2}=4.5 \times 10^{9}$ years.)
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