Chapter 24, Problem 24.117P

Interpretation Introduction

Interpretation:

Decay series of ${}_{92}^{239}U$ has to be given where it first undergo ${\beta }^{-}$ emission, then again emission of ${\beta }^{-}$ and again two consecutive alpha emission. Also natural series that can be started by this sequence has to be given.

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 ${}_{92}^{239}U$ , it first undergo ${\beta }^{-}$ emission, then again emission of ${\beta }^{-}$ and again two consecutive alpha emission.

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 ${\beta }^{-}$ particle from ${}_{92}^{239}U$

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

The given unbalanced nuclear equation is,

  ${}_{92}^{239}U\stackrel{}{\to }{}_{-1}^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{92}^{239}U$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{92}\left(U\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}93\\ \\ Element\text{ with Z = }93\Rightarrow Neptunium\left(Np\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{92}^{239}U\text{ =}{}_{93}^x\text{Np}\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}239\end{array}$

The elemental symbol of the unknown element is ${}_{93}^{239}\text{Np}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{92}^{239}U\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{93}^{239}\text{Np}}$

• Step (2): Emission of ${\beta }^{-}$ particle from ${}_{93}^{239}\text{Np}$

The given unbalanced nuclear equation is,

  ${}_{93}^{239}\text{Np }\stackrel{}{\to }{}_{-1}^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{93}^{239}\text{Np}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{93}\left(Np\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}94\\ \\ Element\text{ with Z = }94\Rightarrow Plutonium\left(Pu\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{93}^{239}\text{Np =}{}_{94}^x\text{Pu}\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}239\end{array}$

The elemental symbol of the unknown element is ${}_{94}^{239}\text{Pu}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{93}^{239}\text{Np }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{94}^{239}\text{Pu}}$

• Step (3): Emission of $\alpha$ particle from ${}_{94}^{239}\text{Pu}$

The given unbalanced nuclear equation is,

  ${}_{94}^{239}\text{Pu}\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{94}^{239}\text{Pu}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{94}\left(\text{Pu}\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}92\\ \\ Element\text{ with Z = }92\Rightarrow Uranium\left(U\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{94}^{239}\text{Pu =}{}_{92}^x\text{U}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{235}\end{array}$

The elemental symbol of the unknown element is ${}_{92}^{235}\text{U}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{94}^{239}\text{Pu}\stackrel{}{\to }{}_{92}^{235}\text{U + }{}_2^4\alpha }$

• Step (4): Emission of $\alpha$ particle from ${}_{92}^{235}\text{U}$

The given unbalanced nuclear equation is,

  ${}_{92}^{233}\text{U}\stackrel{}{\to }?\text{ + }{}_2^4\alpha$

Product formed by $\alpha$ decaying of ${}_{92}^{233}\text{U}$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ \text{92}\left(\text{U}\right)\text{=}\text{x}\text{+ 2}\left(\text{alpha}\right)\\ \text{x}\text{=}90\\ \\ Element\text{ with Z = }90\Rightarrow Thorium\left(Th\right)\\ \\ \text{Sum of Mass number of Reactant = Product}\\ {}_{92}^{233}\text{U =}{}_{90}^x\text{Th}\text{+ 4}\left(\text{alpha}\right)\\ \text{x}\text{=}\text{231}\end{array}$

The elemental symbol of the unknown element is ${}_{90}^{231}\text{Th}$. Therefore, the balanced nuclear equation is,

  $\underset{\_}{{}_{92}^{233}\text{U}\stackrel{}{\to }{}_{90}^{231}\text{Th + }{}_2^4\alpha }$

Therefore, the decay series of ${}_{92}^{239}U$ can be follows as,

  $\begin{array}{c}{}_{92}^{239}U\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{93}^{239}\text{Np}\\ \\ {}_{93}^{239}\text{Np }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{94}^{239}\text{Pu}\\ \\ {}_{94}^{239}\text{Pu}\stackrel{}{\to }{}_{92}^{235}\text{U + }{}_2^4\alpha \\ \\ {}_{92}^{233}\text{U}\stackrel{}{\to }{}_{90}^{231}\text{Th + }{}_2^4\alpha \end{array}$

Sequence of reaction is given below,

${}_{92}^{239}U\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{93}^{239}\text{Np }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{94}^{239}\text{Pu }\stackrel{}{\to }{}_{92}^{235}\text{U + }{}_2^4\alpha \stackrel{}{\to }{}_{90}^{231}\text{Th + }{}_2^4\alpha$

The sequence shown above can begin the decay series of ${}_{92}^{235}U$.