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Chapter 21: Problem 61

Complete and balance the nuclear equations for the following fission or fusion reactions: (a) ${ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{36}^{92} \mathrm{Kr}+{ }_{56}^{141} \mathrm{Zn}+$ (b) ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+$

Short Answer

Expert verified
The short answer is: (a) Completed and balanced fission reaction: \( {}_{92}^{235}U + {}_{0}^{1}n \longrightarrow {}_{36}^{92}Kr + {}_{56}^{141}Zn + {}_{0}^{3}n \) (b) Completed and balanced fusion reaction: \( {}_{1}^{2}H + {}_{1}^{3}H \longrightarrow {}_{2}^{4}He + {}_{0}^{1}n \)

Step by step solution

01

(a) Identify missing particles in the fission equation

: The fission equation is given as: \( {}_{92}^{235}U + {}_{0}^{1}n \longrightarrow {}_{36}^{92}Kr + {}_{56}^{141}Zn + ? \) There is one missing particle in this equation, represented by the question mark (?).
02

(a) Balance atomic number and mass number in the fission equation

: For the atomic number, we have on the left side of the equation: \( 92 + 0 = 92 \) On the right side, we currently have: \( 36 + 56 = 92 \) So, the atomic numbers are balanced. For the mass numbers, on the left side, we have: \( 235 + 1 = 236 \) On the right side, we have: \( 92 + 141 + ? \) So, we need to find the mass number of the missing particle such that it balances the equation: \( 92 + 141 + ? = 236 \) Solving for the missing mass number: \( ? = 236 - 92 - 141 = 3 \)
03

(a) Complete the fission equation

: Now that we know the mass number of the missing particle, we can complete the fission equation. The missing particle is a neutron with a mass number of 3: \( {}_{92}^{235}U + {}_{0}^{1}n \longrightarrow {}_{36}^{92}Kr + {}_{56}^{141}Zn + {}_{0}^{3}n \) The completed and balanced fission equation is: \( {}_{92}^{235}U + {}_{0}^{1}n \longrightarrow {}_{36}^{92}Kr + {}_{56}^{141}Zn + {}_{0}^{3}n \)
04

(b) Identify missing particles in the fusion equation

: The fusion equation is given as: \( {}_{1}^{2}H + {}_{1}^{3}H \longrightarrow {}_{2}^{4}He + ? \) There is one missing particle in this equation, represented by the question mark (?).
05

(b) Balance atomic number and mass number in the fusion equation

: For the atomic number, on the left side of the equation, we have: \( 1 + 1 = 2 \) On the right side, we currently have: \( 2 \) So, the atomic numbers are already balanced. For the mass numbers, on the left side of the equation, we have: \( 2 + 3 = 5 \) On the right side, we have: \( 4 + ? \) So, we need to find the mass number of the missing particle such that it balances the equation: \( 4 + ? = 5 \) Solving for the missing mass number: \( ? = 5 - 4 = 1 \)
06

(b) Complete the fusion equation

: Now that we know the mass number of the missing particle, we can complete the fusion equation. The missing particle is a neutron with a mass number of 1: \( {}_{1}^{2}H + {}_{1}^{3}H \longrightarrow {}_{2}^{4}He + {}_{0}^{1}n \) The completed and balanced fusion equation is: \( {}_{1}^{2}H + {}_{1}^{3}H \longrightarrow {}_{2}^{4}He + {}_{0}^{1}n \)

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Most popular questions from this chapter

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 megawatts? Recall that a watt is \(1 \mathrm{~J} / \mathrm{s}\).

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 grays and sieverts of radiation in 1 yr from such a level of plutonium.

Nuclear scientists have synthesized new elements and isotopes, which are not known in nature using heavy-ion bombardment techniques in high-energy particle accelerators. Complete and balance the following reactions: (a) ${ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \longrightarrow ?+{ }_{2}^{4} \mathrm{He}$ (b) \({ }_{3}^{6} \mathrm{Li}+{ }_{28}^{63} \mathrm{Ni} \longrightarrow\) ? (c) \({ }^{252} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \longrightarrow ?\) (d) ${ }_{92}^{238} \mathrm{U}+{ }_{6}^{12} \mathrm{C} \longrightarrow ?+4{ }_{0}^{1} \mathrm{n}$

Phosphorus- 32 is commonly used in nuclear medicine for the identification of malignant tumors. It decays to sulphur- 32 with a half-life of 14.29 days. If a patient is given 3.5 mg of phosphorus-32, how much phosphorus-32 will remain after 1 month (i.e. 30 days)?

Write balanced nuclear equations for the following transformations: (a) polonium-210 emits alpha particle; (b) neptunium-235 undergoes electron capture; (c) fluorine-18 emits beta particle; (d) carbon-14 decays by beta emission.
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