Chapter 24, Problem 24.53P

(a)

Interpretation Introduction

Interpretation:

Missing species in the given transmutations, ${}^{\text{10}}\text{B}\left(\text{α}\text{,}\text{n}\right)\underset{\_}{}$ and the full nuclear equation has to be written.

Concept Introduction:

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. For example:

  $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

Common particles in radioactive decay and nuclear transformations are mentioned below,

  $\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

• The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.
• The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Explanation of Solution

For the given reaction, short hand notation is ${}^{\text{10}}\text{B}\left(\text{α}\text{,}\text{n}\right)\text{X}$. From the notation it is clear that,

  $\begin{array}{c}\text{Parent}{\text{nucleus : }}^{10}B\\ \text{Projectile : }\alpha \\ \text{Daughter}\text{nucleus : X}\\ \text{Ejectile : n}\end{array}$

The given unbalanced nuclear equation is,

  ${}^{10}B{\text{ + }}_2^4\alpha \stackrel{}{\to }?\text{ + }{}_0^{1}n$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

Missing species is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{5}\left(\text{B}\right)\text{ + 2}\left(\text{alpha}\right)\text{ = x + 0}\left(neutron\right)\\ \text{x = }7\\ \\ Element\text{ with Z = }7\Rightarrow Nitrogen\left(N\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_5^{10}\text{B + 4}\left(\text{alpha}\right){\text{ = }}_7^x\text{N + 1}\left(neutron\right)\\ \text{x = 13}\end{array}$

The elemental symbol of the missing species is ${}_7^{13}N$. Therefore, the balanced nuclear equation is,

  ${}^{10}B{\text{ + }}_2^4\alpha \stackrel{}{\to }{}_7^{13}N{\text{ + }}_0^{1}n$

(b)

Interpretation Introduction

Interpretation:

Missing species in the given transmutations, ${}^{28}Si{\left(\text{d}\text{,}\underset{\_}{}\right)}^{29}P$ and the full nuclear equation has to be written.

Concept Introduction:

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. For example:

  $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

Common particles in radioactive decay and nuclear transformations are mentioned below,

  $\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

• The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.
• The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Explanation of Solution

For the given reaction, short hand notation is ${}^{28}Si{\left(\text{d}\text{,}\text{x}\right)}^{29}P$. From the notation it is clear that,

  $\begin{array}{c}\text{Parent}{\text{nucleus : }}^{28}Si\\ {\text{Projectile : }}^{2}H\\ \text{Daughter}{\text{nucleus : }}^{29}P\\ \text{Ejectile : x}\end{array}$

The given unbalanced nuclear equation is,

  ${}^{28}Si{\text{ + }}^{2}H\stackrel{}{\to }?{\text{ + }}^{29}P$

According to the law of conservation of atomic mass and number, the unknown species can be predicted.

Missing species is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{14}\left(\text{Si}\right)\text{ + 1}\left(\text{deuteron}\right)\text{ = }\left(15\right)P\text{ + }x\\ \text{x = }0\\ \\ \text{Atomic number = 0}\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{14}^{28}\text{Si + 2}\left(\text{deuteron}\right)\text{ = }x{\text{ + }}_{15}^{29}P\\ \text{x = 1}\\ \\ \text{Mass number = 1}\end{array}$

By analyzing the X, atomic number of X is 0 and the atomic mass is 1. So the missing species is neutron, $\left({}_{\text{0}}^{\text{1}}\text{n}\right)$.

Therefore, the balanced nuclear equation is,

  ${}_{14}^{28}Si{\text{ + }}_1^{2}H\stackrel{}{\to }{}_0^{1}n{\text{ + }}_{15}^{29}P$

(c)

Interpretation Introduction

Interpretation:

Missing species in the given transmutations, $\underset{\_}{}{\left(\alpha \text{,}\text{2n}\right)}^{244}Cf$ and the full nuclear equation has to be written.

Concept Introduction:

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. For example:

  $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

Common particles in radioactive decay and nuclear transformations are mentioned below,

  $\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

• The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.
• The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Explanation of Solution

For the given reaction, short hand notation is $\text{X}{\left(\alpha \text{,}\text{2n}\right)}^{244}Cf$. From the notation it is clear that,

  $\begin{array}{c}\text{Parent}{\text{nucleus : }}^{10}B\\ \text{Projectile : }\alpha \\ \text{Daughter}\text{nucleus : X}\\ \text{Ejectile : n}\end{array}$

The given unbalanced nuclear equation is,

  $X{\text{ + }}_2^4\alpha \stackrel{}{\to }{}_{98}^{244}\text{Cf + 2}{}_0^{1}n$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

Missing species is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{x + 2}\left(\text{alpha}\right)\text{ = 98}\left(Cm\right)\text{ + }\left[\text{2}\times \text{0}\left(neutron\right)\right]\\ \text{x = }7\\ \\ Element\text{ with Z = }96\Rightarrow Curium\left(Cm\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{96}^x\text{Cm + 4}\left(\text{alpha}\right){\text{ = }}_{98}^{244}\text{Cf + }\left[\text{2}\times \text{1}\left(neutron\right)\right]\\ \text{x = 242}\end{array}$

The elemental symbol of the missing species is ${}_{96}^{242}Cm$. Therefore, the balanced nuclear equation is,

  ${}_{96}^{242}{\text{Cm + }}_2^4\alpha \stackrel{}{\to }{}_{98}^{244}\text{Cf + 2}{}_0^{1}n$