Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

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

In the decay series of ${}^{232}Th$ , the sequence of particles emitted in each step follows as,

  $\alpha ,{\beta }^{-},{\beta }^{-},\alpha ,\alpha ,\alpha ,\alpha ,{\beta }^{-},{\beta }^{-},\alpha$

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

The elemental symbol of an element: ${}_{\text{Z}}^{\text{A}}\text{X =}{}_{\text{Atomic}\text{number}}^{\text{Mass}\text{number}}\text{X}$

• Step (1): Emission of $\alpha$ particle from ${}^{232}Th$

An alpha particle is emitted and an alpha particle is ${}_2^4\alpha .$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

  ${}_{90}^{232}Th\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}^{232}Th$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{90}\left(\text{Th}\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}88\\ \\ Element\text{ with Z = 88 }\Rightarrow Radium\left(Ra\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{90}^{232}Th\text{ =}{}_{82}^x\text{Ra}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{228}\end{array}$

The elemental symbol of the unknown element is ${}_{88}^{228}\text{Ra}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{90}^{232}Th\stackrel{}{\to }{}_{88}^{228}\text{Ra + }{}_2^4\alpha }$

• Step (2): Emission of ${\beta }^{-}$ particle from ${}_{88}^{228}\text{Ra}$

A ${\beta }^{-}$ particle is emitted and a ${\beta }^{-}$ particle is ${}_{-1}^0\beta$.

The given unbalanced nuclear equation is,

  ${}_{88}^{228}\text{Ra}\stackrel{}{\to }{}_{-1}^0\beta \text{ + ?}$

Product formed by ${\beta }^{-}$ decaying of ${}_{88}^{228}\text{Ra}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{88}\left(Ra\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}89\\ \\ Element\text{ with Z = 89 }\Rightarrow Actinium\left(Ac\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{88}^{228}\text{Ra =}{}_{89}^{x}Ac\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}228\end{array}$

The elemental symbol of the unknown element is ${}_{89}^{228}Ac$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{88}^{228}\text{Ra}\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{89}^{228}Ac}$

• Step (3): Emission of ${\beta }^{-}$ particle from ${}_{89}^{228}Ac$

The given unbalanced nuclear equation is,

  ${}_{89}^{228}Ac\stackrel{}{\to }{}_{-1}^0\beta \text{ + ?}$

Product formed by ${\beta }^{-}$ decaying of ${}_{89}^{228}Ac$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{89}\left(Ac\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}90\\ \\ Element\text{ with Z = 90 }\Rightarrow Thorium\left(Th\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{89}^{228}Ac\text{ =}{}_{90}^{x}Th\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}228\end{array}$

The elemental symbol of the unknown element is ${}_{90}^{228}Th$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{89}^{228}Ac\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{90}^{228}Th}$

• Step (4): Emission of $\alpha$ particle from ${}_{90}^{228}Th$

The given unbalanced nuclear equation is,

  ${}_{90}^{228}Th\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{90}^{228}Th$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{90}\left(Th\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}88\\ \\ Element\text{ with Z = 88 }\Rightarrow Radium\left(Ra\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{90}^{228}Th\text{ =}{}_{88}^x\text{Ra}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{224}\end{array}$

The elemental symbol of the unknown element is ${}_{88}^{224}\text{Ra}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{90}^{228}Th\stackrel{}{\to }{}_{88}^{224}\text{Ra+ }{}_2^4\alpha }$

• Step (5): Emission of $\alpha$ particle from ${}_{88}^{224}\text{Ra}$

The given unbalanced nuclear equation is,

  ${}_{88}^{224}\text{Ra}\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{88}^{224}\text{Ra}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{88}\left(\text{Ra}\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}86\\ \\ Element\text{ with Z = 86 }\Rightarrow Radon\left(Rn\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{88}^{224}\text{Ra =}{}_{86}^x\text{Rn}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{220}\end{array}$

The elemental symbol of the unknown element is ${}_{86}^{220}\text{Rn}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{88}^{224}\text{Ra}\stackrel{}{\to }{}_{86}^{220}\text{Rn + }{}_2^4\alpha }$

• Step (6): Emission of $\alpha$ particle from ${}_{86}^{220}\text{Rn}$

The given unbalanced nuclear equation is,

  ${}_{86}^{220}\text{Rn}\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{86}^{220}\text{Rn}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{86}\left(Rn\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}84\\ \\ Element\text{ with Z = 84 }\Rightarrow Polonium\left(Po\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{86}^{220}\text{Rn =}{}_{84}^x\text{Po}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{216}\end{array}$

The elemental symbol of the unknown element is ${}_{84}^{216}\text{Po}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{86}^{220}\text{Rn}\stackrel{}{\to }{}_{84}^{216}\text{Po + }{}_2^4\alpha }$

• Step (7): Emission of $\alpha$ particle from ${}_{84}^{216}\text{Po}$

The given unbalanced nuclear equation is,

  ${}_{84}^{216}\text{Po}\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{84}^{216}\text{Po}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{84}\left(\text{Po}\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}82\\ \\ Element\text{ with Z = 82 }\Rightarrow Lead\left(Pb\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{84}^{216}\text{Po =}{}_{82}^x\text{Pb}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{212}\end{array}$

The elemental symbol of the unknown element is ${}_{82}^{212}\text{Pb}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{84}^{216}\text{Po}\stackrel{}{\to }{}_{82}^{212}\text{Pb + }{}_2^4\alpha }$

• Step (8): Emission of ${\beta }^{-}$ particle from ${}_{82}^{212}\text{Pb}$

The given unbalanced nuclear equation is,

  ${}_{82}^{212}\text{Pb}\stackrel{}{\to }{}_{-1}^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{82}^{212}\text{Pb}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{82}\left(Pb\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}83\\ \\ Element\text{ with Z = }83\Rightarrow Bismuth\left(Bi\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{82}^{212}\text{Pb =}{}_{83}^x\text{Bi}\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}212\end{array}$

The elemental symbol of the unknown element is ${}_{83}^{212}\text{Bi}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{82}^{209}\text{Pb }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{83}^{212}\text{Bi}}$

• Step (9): Emission of ${\beta }^{-}$ particle from ${}_{83}^{212}\text{Bi}$

The given unbalanced nuclear equation is,

  ${}_{83}^{212}\text{Bi }\stackrel{}{\to }{}_{-1}^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{83}^{212}\text{Bi}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{83}\left(Bi\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}84\\ \\ Element\text{ with Z = }84\Rightarrow Polonium\left(Po\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{83}^{212}\text{Bi =}{}_{84}^x\text{Po}\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}212\end{array}$

The elemental symbol of the unknown element is ${}_{84}^{212}\text{Po}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{83}^{212}\text{Bi }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{84}^{212}\text{Po}}$

• Step (10): Emission of $\alpha$ particle from ${}_{84}^{212}\text{Po}$

The given unbalanced nuclear equation is,

  ${}_{84}^{212}\text{Po}\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{84}^{212}\text{Po}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{84}\left(\text{Po}\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}82\\ \\ Element\text{ with Z = 82 }\Rightarrow Lead\left(Pb\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{84}^{212}\text{Po =}{}_{82}^x\text{Pb}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{208}\end{array}$

The elemental symbol of the unknown element is ${}_{82}^{208}\text{Pb}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{84}^{212}\text{Po}\stackrel{}{\to }{}_{82}^{208}\text{Pb + }{}_2^4\alpha }$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

  $\begin{array}{l}{}_{90}^{232}Th\stackrel{}{\to }{}_{88}^{228}\text{Ra + }{}_2^4\alpha \\ \\ {}_{88}^{228}\text{Ra}\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{89}^{228}Ac\\ \\ {}_{89}^{228}Ac\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{90}^{228}Th\\ \\ {}_{90}^{228}Th\stackrel{}{\to }{}_{88}^{224}\text{Ra+ }{}_2^4\alpha \\ \\ {}_{88}^{224}\text{Ra}\stackrel{}{\to }{}_{86}^{220}\text{Rn + }{}_2^4\alpha \\ \\ {}_{86}^{220}\text{Rn}\stackrel{}{\to }{}_{84}^{216}\text{Po + }{}_2^4\alpha \\ \\ {}_{84}^{216}\text{Po}\stackrel{}{\to }{}_{82}^{212}\text{Pb + }{}_2^4\alpha \\ \\ {}_{82}^{209}\text{Pb }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{83}^{212}\text{Bi}\\ \\ {}_{83}^{212}\text{Bi }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{84}^{212}\text{Po}\\ \\ {}_{84}^{212}\text{Po}\stackrel{}{\to }{}_{82}^{208}\text{Pb + }{}_2^4\alpha \end{array}$