Chapter 24, Problem 24.107P

(a)

Interpretation Introduction

Interpretation:

The overall equation in shorthand notation has to be written for the given reaction starting with the stable isotope before neutron activation.

Concept Introduction:

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.

In NAA (neutron activation analysis), non-radioactive sample is bombarded with the neutrons where a small part of its atoms get converted to radioisotopes. These radioisotopes get decayed by emitting different radiations for each isotope.

Explanation of Solution

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

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

Predict the product formed when Vanadium-51 bombarded with neutron:

The given unbalanced nuclear equation is,

  ${}_{23}^{51}{\text{V + }}_0^1\text{n }\stackrel{}{\to }?$

Product formed is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ 23\left(V\right)\text{ + 0}\left(neutron\right)\text{= x}\\ \text{x}\text{ =}23\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{23}^{51}\text{V + 1}\left(neutron\right)\text{ =}{}_{23}^{x}V\\ \text{x = 52}\end{array}$

The elemental symbol of the product formed is ${}_{23}^{52}V$. Therefore, the balanced nuclear equation is ${}_{23}^{51}{\text{V + }}_0^1\text{n }\stackrel{}{\to }{}_{23}^{52}V.$

Predict the product formed after beta emission of Vanadium-52:

The given unbalanced nuclear equation is,

  ${}_{23}^{52}V\stackrel{}{\to }{}_{-1}^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{23}^{52}V$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ 23\left(Cr\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}24\\ \\ Element\text{ with Z = }24\Rightarrow Chromium\left(Cr\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{23}^{52}V\text{ =}{}_{24}^{x}Cr\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}52\end{array}$

The elemental symbol of the unknown element is ${}_{24}^{52}Cr$. Therefore, the balanced nuclear equation is ${}_{23}^{52}V\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{24}^{52}Cr.$

Therefore,

The overall equation in shorthand notation for the reaction is shown below,

  $\begin{array}{l}{}_{23}^{51}{\text{V + }}_0^1\text{n }\stackrel{}{\to }{}_{23}^{52}V\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{24}^{52}Cr\\ \\ {}_{23}^{51}\text{V}{\left(n,\beta \right)}_{24}^{52}Cr\end{array}$

(b)

Interpretation Introduction

Interpretation:

The overall equation in shorthand notation has to be written for the given reaction starting with the stable isotope before neutron activation.

Concept Introduction:

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.

In NAA (neutron activation analysis), non-radioactive sample is bombarded with the neutrons where a small part of its atoms get converted to radioisotopes. These radioisotopes get decayed by emitting different radiations for each isotope.

Explanation of Solution

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

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

Predict the product formed when Copper-63 bombarded with neutron:

The given unbalanced nuclear equation is,

  ${}_{29}^{63}{\text{Cu + }}_0^1\text{n }\stackrel{}{\to }?$

Product formed is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ 29\left(Cu\right)\text{ + 0}\left(neutron\right)\text{= x}\\ \text{x}\text{ =}29\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{29}^{63}\text{Cu + 1}\left(neutron\right){\text{ = }}_{29}^x\text{Cu }\\ \text{x = 64}\end{array}$

The elemental symbol of the product formed is ${}_{29}^{64}\text{Cu }$. Therefore, the balanced nuclear equation is ${}_{29}^{63}{\text{Cu + }}_0^1\text{n }\stackrel{}{\to }{}_{29}^{64}\text{Cu}\text{.}$

Predict the product formed after positron emission of Copper-64:

The given unbalanced nuclear equation is,

  ${}_{29}^{64}\text{Cu }\stackrel{}{\to }{}_1^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{23}^{52}V$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ 29\left(Cu\right)\text{ = x+ }\text{1}\left({\beta }^{+}\right)\\ \text{x}\text{ =}28\\ \\ Element\text{ with Z = }28\Rightarrow Nickel\left(Ni\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{29}^{64}\text{Cu =}{}_{28}^{x}Ni\text{ + 0}\left({\beta }^{+}\right)\\ \text{x = 64}\end{array}$

The elemental symbol of the unknown element is ${}_{28}^{64}Ni$. Therefore, the balanced nuclear equation is ${}_{29}^{64}\text{Cu }\stackrel{}{\to }{}_1^0\beta {\text{ + }}_{28}^{64}Ni.$

Therefore,

The overall equation in shorthand notation for the reaction is shown below,

  $\begin{array}{l}{}_{29}^{63}{\text{Cu + }}_0^1\text{n }\stackrel{}{\to }{}_{29}^{64}\text{Cu }\stackrel{}{\to }{}_1^0\beta {\text{ + }}_{28}^{64}Ni.\\ \\ {}_{29}^{63}\text{Cu}{\left(n,{\beta }^{+}\right)}_{28}^{64}Ni\end{array}$

(c)

Interpretation Introduction

Interpretation:

The overall equation in shorthand notation has to be written for the given reaction starting with the stable isotope before neutron activation.

Concept Introduction:

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.

In NAA (neutron activation analysis), non-radioactive sample is bombarded with the neutrons where a small part of its atoms get converted to radioisotopes. These radioisotopes get decayed by emitting different radiations for each isotope.

Explanation of Solution

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

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

Predict the product formed when Aluminum-27 bombarded with neutron:

The given unbalanced nuclear equation is,

  ${}_{13}^{27}{\text{Al + }}_0^1\text{n }\stackrel{}{\to }?$

Product formed is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ 13\left(Al\right)\text{ + 0}\left(neutron\right)\text{= x}\\ \text{x}\text{ =}13\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{13}^{27}\text{Al + 1}\left(neutron\right){\text{ = }}_{13}^x\text{Al}\\ \text{x = 28}\end{array}$

The elemental symbol of the product formed is ${}_{13}^{28}\text{Al}$. Therefore, the balanced nuclear equation is ${}_{13}^{27}{\text{Al + }}_0^1\text{n }\stackrel{}{\to }{}_{13}^{28}\text{Al}.$

Predict the product formed after beta emission of Aluminum-28:

The given unbalanced nuclear equation is,

  ${}_{13}^{28}\text{Al }\stackrel{}{\to }{}_{-1}^0\beta \text{ + }?$

Product formed by ${\beta }^{-}$ decaying of ${}_{23}^{52}V$ is determined as follows,

  $\begin{array}{l}\text{Sum of Atomic number of Reactant = Product}\\ 23\left(Cr\right)\text{ = x+ }-\text{1}\left({\beta }^{-}\right)\\ \text{x}\text{ =}24\\ \\ Element\text{ with Z = }24\Rightarrow Chromium\left(Cr\right)\\ \\ \text{Sum of Atomic number of Reactant = Product}\\ {}_{13}^{28}\text{Al =}{}_{14}^{x}Si\text{ + 0}\left({\beta }^{-}\right)\\ \text{x}\text{=}28\end{array}$

The elemental symbol of the unknown element is ${}_{14}^{28}Si$. Therefore, the balanced nuclear equation is ${}_{13}^{28}\text{Al }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{14}^{28}Si.$

Therefore,

The overall equation in shorthand notation for the reaction is shown below,

  $\begin{array}{l}{}_{13}^{27}{\text{Al + }}_0^1\text{n }{\stackrel{}{\to }}_{13}^{28}\text{Al }\stackrel{}{\to }{}_{-1}^0\beta {\text{ + }}_{14}^{28}Si\\ \\ {}_{13}^{27}\text{Al}{\left(n,\beta \right)}_{14}^{28}Si\end{array}$