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Chapter 21: Problem 30

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{17}^{14} \mathrm{N}+_{2}^{4} \mathrm{He} \longrightarrow ? +_{1}^{1} \mathrm{H}} \\ {\text { (b) }_{19}^{40} \mathrm{K}+_{1}^{0} \mathrm{e} \ \mathrm{(orbital \ electron) \longrightarrow ?}}\\\ {\text { (c) }_{}{} \mathrm{?}+_{2}^{4} \mathrm{He} \longrightarrow_{14}^{30} \mathrm{Si} +_{1}^{1} \mathrm{H}}\\\ {\text { (d) }_{26}^{58} \mathrm{Fe} +2 _{0}^{1} \mathrm{n} \longrightarrow_{27}^{60} \mathrm{Co}+?}\\\ {\text { (e) }_{92}^{235} \mathrm{U}\longrightarrow+_{0}^{1} n \longrightarrow_{54}^{135} \mathrm{Xe}+2_{0}^{1} \mathrm{n}+?} \end{array}\end{equation}

Short Answer

Expert verified
The short answer for the balanced nuclear equations is: (a) \( \ _{17}^{14}N\ +\ _{2}^{4}He\ \longrightarrow\ _{18}^{17}O\ +\ _{1}^{1}H\) (b) \( \ _{19}^{40}K\ +\ _{1}^{0}e\ \longrightarrow\ _{19}^{40}Ca\) (c) \( \ _{13}^{27}Al\ +\ _{2}^{4}He\ \longrightarrow\ _{14}^{30}Si\ +\ _{1}^{1}H\) (d) \( \ _{26}^{58}Fe\ +\ _{0}^{1}n\ \longrightarrow\ _{27}^{60}Co\ +\ _{0}^{-1}e^-\) (e) \( \ _{92}^{235}U\ +\ _{0}^{1}n\ \longrightarrow\ _{54}^{135}Xe\ +\ _{0}^{1}n\ +\ _{38}^{99}Sr\)

Step by step solution

01

Determine the missing particle's atomic and mass numbers

First, we need to find out the atomic number (Z) and mass number (A) of the unknown particle (?). We will use the fact that in a balanced nuclear equation, the total atomic numbers (Z) and mass numbers (A) are conserved.
02

Write down the equation for atomic number conservation

\( Z_1 + Z_2 = Z_3 + Z_4 \Rightarrow 17 + 2 = Z_3 + 1 \) Solving for Z3: \( Z_3 = 18\)
03

Write down the equation for mass number conservation

\( A_1 + A_2 = A_3 + A_4 \Rightarrow 14 + 4 = A_3 + 1 \) Solving for A3: \( A_3 = 17 \) The missing particle is \(_{18}^{17}O\). The balanced equation is: \(_{17}^{14}N\ +\ _{2}^{4}He\ \longrightarrow\ _{18}^{17}O\ +\ _{1}^{1}H\) (b) Balancing the nuclear equation:
04

Determine the missing particle's atomic and mass numbers

Use the conservation laws for atomic number and mass number to identify the unknown particle.
05

Write down the equation for atomic number conservation

\( Z_1 + Z_2 = Z_3 \Rightarrow 19 = Z_3 \)
06

Write down the equation for mass number conservation

\( A_1 + A_2 = A_3 \Rightarrow 40 = A_3 \) The missing particle is \(_{19}^{40}Ca\). The balanced equation is: \(_{19}^{40}K\ +\ _{1}^{0}e\ \longrightarrow\ _{19}^{40}Ca\) (c) Balancing the nuclear equation:
07

Determine the missing particle's atomic and mass numbers

Use the conservation laws for atomic number and mass number to identify the unknown particle.
08

Write down the equation for atomic number conservation

\( Z_1 + Z_2 = Z_3 + Z_4 \Rightarrow Z_1 + 2 = 14 + 1 \) Solving for Z1: \( Z_1 = 13\)
09

Write down the equation for mass number conservation

\( A_1 + A_2 = A_3 + A_4 \Rightarrow A_1 + 4 = 30 + 1 \) Solving for A1: \( A_1 = 27 \) The missing particle is \(_{13}^{27}Al\). The balanced equation is: \(_{13}^{27}Al\ +\ _{2}^{4}He\ \longrightarrow\ _{14}^{30}Si\ +\ _{1}^{1}H\) (d) Balancing the nuclear equation:
10

Determine the missing particle's atomic and mass numbers

Use the conservation laws for atomic number and mass number to identify the unknown particle.
11

Write down the equation for atomic number conservation

\( Z_1 + Z_2 = Z_3 \Rightarrow 26 + 0 = 27\)
12

Write down the equation for mass number conservation

\( A_1 + A_2 = A_3 + A_4 \Rightarrow 58 + 1 = 60 + A_4 \) Solving for A4: \( A_4 = -1 \) The missing particle is \(_{0}^{-1}e^-\). The balanced equation is: \(_{26}^{58}Fe\ +\ _{0}^{1}n\ \longrightarrow\ _{27}^{60}Co\ +\ _{0}^{-1}e^-\) (e) Balancing the nuclear equation:
13

Determine the missing particle's atomic and mass numbers

Use the conservation laws for atomic number and mass number to identify the unknown particle.
14

Write down the equation for atomic number conservation

\( Z_1 + Z_2 = Z_3 + Z_4 + Z_5 \Rightarrow 92 = 54 + 0 + Z_5 \) Solving for Z5: \( Z_5 = 38\)
15

Write down the equation for mass number conservation

\( A_1 + A_2 = A_3 + A_4 + A_5 \Rightarrow 235 + 1 = 135 + 1 + A_5 \) Solving for A5: \( A_5 = 99 \) The missing particle is \(_{38}^{99}Sr\). The balanced equation is: \(_{92}^{235}U\ +\ _{0}^{1}n\ \longrightarrow\ _{54}^{135}Xe\ +\ _{0}^{1}n\ +\ _{38}^{99}Sr\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Number Conservation
In a nuclear reaction, the atomic number, which represents the number of protons in an atom, must remain the same both before and after the reaction. This is a crucial part of balancing nuclear equations. The atomic number is denoted as "Z".

During any nuclear process, nuclear stability is preserved by ensuring that the total atomic numbers on the left side of the equation equal those on the right side. This principle allows us to solve for unknown particles in nuclear equations by setting up simple equations that sum these atomic numbers.
  • For example, in nuclear equation (a), the atomic numbers must add up to the same total. Thus, for the reaction of N and He to produce an unknown and H:
    \( Z_1 + Z_2 = Z_3 + Z_4 \rightarrow 17 + 2 = 18 + 1 \)
  • This calculation results in \( Z_3 = 18 \), helping us identify the missing particle's atomic number.
Mass Number Conservation
The mass number in a nuclear equation, denoted "A", is the total number of protons and neutrons in an atomic nucleus. Just like the atomic number, the total mass number must be conserved in nuclear reactions. This means the sum of mass numbers on the reactant side equals the sum on the product side. This principle helps maintain the identity of atomic species involved.

For example, each side of the nuclear equation should balance with the same total mass. This ensures no loss or creation of matter according to classical physics principles.
  • In example (b), consider the conservation of mass numbers:
    \( A_1 + A_2 = A_3 \rightarrow 40 = 40 \)
  • In this case, the conservation of mass number implies no additional particles are required for the reaction from K to Ca.
Balancing Nuclear Equations
Balancing nuclear equations involves ensuring both atomic and mass numbers balance across the equation. It supports the laws of conservation and allows chemical equations to adhere to physical realities.

To balance nuclear equations, one has to write down both atomic and mass number conservation equations after analyzing the given nuclear reaction. By solving these equations, the missing particles can be determined, ensuring the total charge and mass sums match both reactants and products.
  • This method often leads us to identify new isotopes, beta particles, or even positrons based on what's required to balance the equation.
  • In exercise (c), by balancing atomic numbers and mass numbers, we ascertain the unknown particle is \(_{13}^{27}Al\) because:
    \( Z_1 + 2 = 14 + 1 \rightarrow Z_1 = 13 \) and \( A_1 + 4 = 30 + 1 \rightarrow A_1 = 27 \).

This process exemplifies how nuclear equations reveal the nature of nuclear transformations, ensuring every involved atomic species is appropriately accounted for.

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Most popular questions from this chapter

Naturally found uranium consists of 99.274\(\%^{238} \mathrm{U}\) \(0.720 \%^{233} \mathrm{U},\) and 0.006\(\%^{233} \mathrm{U}\) As we have seen, \(^{235} \mathrm{U}\) is the isotope that can undergo a nuclear chain reaction. Most of the \(^{255}\) U used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, UF \(_{6}(g) .\) (a) What is the mass of UF \(_{6}\) in a 30.0 -L vessel of UF \(_{6}\) at a pressure of 695 torr at 350 \(\mathrm{K} ?\) (b) What is the mass of \(^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the \(\mathrm{UF}_{6}\) is diffused through a porous barrier and that the change in the ratio of of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) in the diffused gas can be described by Equation 10.23. What is the mass of \(^{235} \mathrm{U}\) in a sample of the diffused gas analogous to that in part (a)? (d) After one more cycle of gaseous diffusion, what is the percentage of \(^{235} \mathrm{UF}_{6}\) in the sample?

Which are classified as ionizing radiation: X rays, alpha particles, microwaves from a cell phone, and gamma rays?

A sample of an alpha emitter having an activity of 0.18 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?

In 2010, a team of scientists from Russia and the United States reported creation of the first atom of element 117, which is named tennessine, and whose symbol is Ts. The synthesis involved the collision of a target of \(_{97}^{249} \mathrm{Bk}\) with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms \(_{97}^{249} \mathrm{Bk} :\) $$_{97}^{249} \mathrm{Bk}+\mathrm{Q} \longrightarrow \mathrm{Z} \longrightarrow_{117 \mathrm{Ts}}^{294 \mathrm{Ts}}+3_{0}^{1} \mathrm{n}$$ (a) What are the identities of isotopes Q and Z? (b) Isotope Q is unusual in that it is very long-lived (its half-life is on the order of 1019 yr) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Collision of ions of isotope Q with a target was also used to produce the first atoms of livermorium, Lv. The initial product of this collision was \(_{116}^{296} \mathrm{Zn}\). What was the target isotope with which Q collided in this experiment?

Indicate the number of protons and neutrons in the following nuclei: \((\mathbf{a}) _{24}^{56} \mathrm{Cr},(\mathbf{b})^{193} \mathrm{Tl},(\mathbf{c})\) argon-\(38.\)
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