Chapter 32, Problem 32.1QAP

Identify X in each of the following nuclear reactions:
(a) $\text{Z}\text{n}+n\to \text{N}\text{i}+\text{X}$
(b) $\text{P}\to \text{S}\text{i}+\text{X}$
(c) $\text{P}\text{b}\to \text{B}\text{i}+\text{X}$
(d) $\text{U}+n\to 4(\text{n}\text{) }+\text{Z}\text{n+X}$
(e) $\text{T}e+\text{H}\to \text{I}\text{+X}$
(f) $\text{C}\text{u}+\text{X}\to \text{N}\text{i}$

Interpretation Introduction

Interpretation:

The value of X is to be determined in the given nuclear reaction

  ${}_{30}^{68}Zn{+}_0^{1}n{\to }_{28}^{65}Ni+X$

Concept Introduction :

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or of subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

Answer to Problem 32.1QAP

The reaction is −

  ${}_{30}^{68}Zn{+}_0^{1}n{\to }_{28}^{65}Ni{+}_2^{4}He$ , where X is ${}_2^{4}He$ .

Explanation of Solution

The given reaction is −

  ${}_{30}^{68}Zn{+}_0^{1}n{\to }_{28}^{65}Ni+X$

The above reaction shows the alpha decay. The alpha decay of element ${}_{30}^{68}Zn$ results in the formation of ${}_{28}^{65}Ni$ and ${}_2^{4}He$ . Therefore, for the given reaction the value of X is ${}_2^{4}He$ . Hence the correct nuclear reaction is given as-

  ${}_{30}^{68}Zn{+}_0^{1}n{\to }_{28}^{65}Ni{+}_2^{4}He$

Interpretation Introduction

Interpretation:

The value of X is to be determined in the given nuclear reaction

  ${}_{15}^{30}P{\to }_{14}^{30}Si+X$

Concept Introduction :

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or of subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

Answer to Problem 32.1QAP

The reaction is −

  ${}_{15}^{30}P{\to }_{14}^{30}Si{+}_1^{0}n$ , where X is ${}_1^{0}n$ .

Explanation of Solution

The given reaction is −

  ${}_{15}^{30}P{\to }_{14}^{30}Si+X$

The above reaction shows the beta positive decay. The beta positive decay of given reaction results in the formation of ${}_{14}^{30}S$ and ${}_1^{0}n$ . Therefore, for the given reaction the value of X is ${}_1^{0}n$ means ${\beta }^{+}$ . Hence the correct nuclear reaction is given as-

  ${}_{15}^{30}P{\to }_{14}^{30}Si{+}_1^{0}n$

Interpretation Introduction

Interpretation:

The value of X is to be determined in the given nuclear reaction

  ${}_{82}^{214}Pb{\to }_{83}^{214}Bi+X$

Concept Introduction :

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or of subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

Answer to Problem 32.1QAP

The reaction is −

  ${}_{82}^{214}Pb{\to }_{83}^{214}Bi{+}_{-1}^{0}n$, where X is ${}_{-1}^{0}n$ .

Explanation of Solution

The given reaction is −

  ${}_{82}^{214}Pb{\to }_{83}^{214}Bi+X$

The above reaction shows the beta decay. The beta decay of given reaction results in the formation of ${}_{83}^{214}Bi$ and ${}_{-1}^{0}n$ . Therefore, for the given reaction the value of X is ${}_{-1}^{0}n$ means ${\beta }^{-}$ . Hence, the correct nuclear reaction is given as-

  ${}_{82}^{214}Pb{\to }_{83}^{214}Bi{+}_{-1}^{0}n$

Interpretation Introduction

Interpretation:

The value of X is to be determined in the given nuclear reaction.

  ${}_{92}^{235}U{+}_0^{1}n\to 4{(}_0^{1}n){+}_{30}^{72}Zn+X$

Concept Introduction:

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or of subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

Answer to Problem 32.1QAP

The reaction is −

  ${}_{92}^{235}U{+}_0^{1}n\to 4{(}_0^{1}n){+}_{30}^{72}Zn{+}_{62}^{160}Sm$ , where X is ${}_{62}^{160}Sm$

Explanation of Solution

The given reaction is −

  ${}_{92}^{235}U{+}_0^{1}n\to 4{(}_0^{1}n){+}_{30}^{72}Zn+X$

The above reaction shows the alpha decay. The alpha decay of element ${}_{92}^{235}U$ results in the formation of ${}_{30}^{72}Zn$ and ${}_{62}^{160}Sm$ . Therefore, for the given reaction the value of X is ${}_{62}^{160}Sm$ . Hence the correct nuclear reaction is given as-

  ${}_{92}^{235}U{+}_0^{1}n\to 4{(}_0^{1}n){+}_{30}^{72}Zn{+}_{62}^{160}Sm$

Interpretation Introduction

Interpretation:

The value of X is to be determined in the given nuclear reaction

  ${}_{52}^{130}Te{+}_1^{2}H{\to }_{53}^{131}I+X$

Concept Introduction:

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or of subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

Answer to Problem 32.1QAP

The reaction is −

  ${}_{52}^{130}Te{+}_1^{2}H{\to }_{53}^{131}I{+}_0^{1}n$ , where X is ${}_0^{1}n$ .

Explanation of Solution

The given reaction is −

  ${}_{52}^{130}Te{+}_1^{2}H{\to }_{53}^{131}I+X$

The above reaction will be balanced when one is added to the mass number. Therefore one neutron with mass number one is added to the reaction, so the value of X is ${}_0^{1}n$ means neutron. Hence the reaction is −

  ${}_{52}^{130}Te{+}_1^{2}H{\to }_{53}^{131}I{+}_0^{1}n$

Interpretation Introduction

Interpretation:

The value of X is to be determined in the given nuclear reaction

  ${}_{29}^{64}Cu+X{\to }_{28}^{64}Ni$

Concept Introduction:

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or of subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

Answer to Problem 32.1QAP

The reaction is −

  ${}_{29}^{64}Cu{+}_1^0{e}^{-}{\to }_{28}^{64}Ni$, where X is ${}_1^0{e}^{-}$

Explanation of Solution

The given reaction is −

  ${}_{29}^{64}Cu+X{\to }_{28}^{64}Ni$

The above reaction will be balanced when one is added to the atomic number. Therefore, one electron with atomic number one will be added to the reaction so the value of X is ${}_1^0{e}^{-}$ . Hence the reaction is −

  ${}_{29}^{64}Cu{+}_1^0{e}^{-}{\to }_{28}^{64}Ni$