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Chapter 21: Problem 29

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{98}^{252} \mathrm{Cf}+_{5}^{10} \mathrm{B} \longrightarrow 3_{0}^{1} \mathrm{n}+?} \\\ {\text { (b) }_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+?}\\\ {\text { (c) }_{1}^{1} \mathrm{H}+_{5}^{11} \mathrm{B} \longrightarrow 3?} \\ {\text { (d) }_{53}^{122} \mathrm{I}\longrightarrow_{54}^{122} \mathrm{Xe}+?}\\\ {\text { (e) }_{26}^{59} \mathrm{Fe}\longrightarrow_{-1}^{0} \mathrm{e}+?}\end{array}\end{equation}

Short Answer

Expert verified
Here are the short answers to the given nuclear reactions: (a) \(\ce{_{98}^{252}Cf + _5^{10}B -> 3_0^1n + _{103}^{259}Lr}\) (b) \(\ce{_{1}^{2}H + _2^{3}He -> _2^{4}He + _1^1H}\) (c) \(\ce{_{1}^{1}H + _5^{11}B -> 3_{2}^{4}He}\) (d) \(\ce{_{53}^{122}I -> _{54}^{122}Xe + _{-1}^0e}\) (e) \(\ce{_{26}^{59}Fe -> _{-1}^{0}e + _{27}^{59}Co}\)

Step by step solution

01

(a) Reaction: Cf + B -> 3n + ?

First, let's write down the given reaction: \[\ce{_{98}^{252}Cf + _5^{10}B -> 3_0^1n + ?}\] To find the missing particle, we need to ensure the conservation of mass number and atomic number: Total mass number on left side = 252 + 10 = 262 Total atomic number on left side = 98 + 5 = 103 Now, there are 3 neutrons on the right side: Mass number accounted for on right side = 3 * 1 = 3 Atomic number accounted for on right side = 0 (Since neutrons have an atomic number of 0) For conservation of mass number: 262 - 3 = 259 For conservation of atomic number: 103 - 0 = 103 So, the missing particle is \(\ce{_{103}^{259}Lr}\). The balanced equation is: \[\ce{_{98}^{252}Cf + _5^{10}B -> 3_0^1n + _{103}^{259}Lr}\]
02

(b) Reaction: H + He -> He + ?

Write down the given reaction: \[\ce{_{1}^{2}H + _2^{3}He -> _2^{4}He + ?}\] To find the missing particle, we need to ensure the conservation of mass number and atomic number: Total mass number on left side = 2 + 3 = 5 Total atomic number on left side = 1 + 2 = 3 On the right side we have: Mass number accounted for = 4 Atomic number accounted for = 2 For conservation of mass number: 5 - 4 = 1 For conservation of atomic number: 3 - 2 = 1 So, the missing particle is \(\ce{_{1}^{1}H}\) (a proton). The balanced equation is: \[\ce{_{1}^{2}H + _2^{3}He -> _2^{4}He + _1^1H}\]
03

(c) Reaction: H + B -> 3?

Write down the given reaction: \[\ce{_{1}^{1}H + _5^{11}B -> 3?}\] Total mass number on left side = 1 + 11 = 12 Total atomic number on left side = 1 + 5 = 6 Since there are 3 identical particles on the right side, let's denote the mass number of each as m and atomic number as a: For conservation of mass number: 12 = 3 * m For conservation of atomic number: 6 = 3 * a Solving these equations, we get m = 4 and a = 2. Thus, the missing particle is \(\ce{_{2}^{4}He}\). The balanced equation is: \[\ce{_{1}^{1}H + _5^{11}B -> 3_{2}^{4}He}\]
04

(d) Reaction: I -> Xe + ?

Write down the given reaction: \[\ce{_{53}^{122}I -> _{54}^{122}Xe + ?}\] Total mass number on left side = 122 Total atomic number on left side = 53 On the right side, we have: Mass number accounted for = 122 Atomic number accounted for = 54 For conservation of mass number: 122 - 122 = 0 For conservation of atomic number: 53 - 54 = -1 So, the missing particle is \(\ce{_{-1}^{0}e}\) (an electron). The balanced equation is: \[\ce{_{53}^{122}I -> _{54}^{122}Xe + _{-1}^0e}\]
05

(e) Reaction: Fe -> e + ?

Write down the given reaction: \[\ce{_{26}^{59}Fe -> _{-1}^{0}e + ?}\] Total mass number on left side = 59 Total atomic number on left side = 26 On the right side, we have: Mass number accounted for = 0 Atomic number accounted for = -1 For conservation of mass number: 59 - 0 = 59 For conservation of atomic number: 26 - (-1) = 27 So, the missing particle is \(\ce{_{27}^{59}Co}\). The balanced equation is: \[\ce{_{26}^{59}Fe -> _{-1}^{0}e + _{27}^{59}Co}\]

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Most popular questions from this chapter

Chlorine has two stable nuclides, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl} .\) In contrast, \(^{36} \mathrm{Cl}\) is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \(^{36} \mathrm{Cl} ?\) (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \(^{36} \mathrm{C}\) is less stable than either \(^{35}\mathrm{Cl}\) or \(^{37} \mathrm{Cl}\).

A wooden artifact from a Chinese temple has a \(^{14} \mathrm{C}\) cactivity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the half-life for \(^{14} \mathrm{C}\) decay, 5715 yr, determine the age of the artifact.

Which of the following statements best explains why alpha emission is relatively common, but proton emission is extremely rare? \begin{equation}\begin{array}{l}{\text { (a) Alpha particles are very stable because of magic numbers }} \\ \quad {\text { of protons and neutrons. }} \\\ {\text { (b) Alpha particles occur in the nucleus. }} \\ {\text { (c) Alpha particles are the nuclei of an inert gas. }} \\ {\text { (d) An alpha particle has a higher charge than a proton. }}\end{array}\end{equation}

Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments: \begin{equation} (a) \stackrel{6}{3} \mathrm{Li}+\stackrel{56}{28} \mathrm{Ni} \longrightarrow ? \end{equation}\begin{equation}(b) \stackrel{40}{20} \mathrm{Ca}+\stackrel {248}{96} \mathrm{Cm} \longrightarrow \stackrel{147}{62} \mathrm{Sm} + ? \end{equation}\begin{equation}(c) \stackrel{88}{38} \mathrm{Sr}+\stackrel{84}{36} \mathrm{Kr} \longrightarrow \stackrel{116}{46} \mathrm{Pd} + ?\end{equation}\begin{equation} (d)\stackrel{40}{20} \mathrm{Ca}+\stackrel{238}{92} \mathrm{U} \longrightarrow \stackrel{70}{30} \mathrm{Zn}+4 \stackrel{1}{0}\mathrm{n}+2 ?\end{equation}

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?
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