Chapter 18, Problem 11QAP

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The production of Pa-233 through alpha emission.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 11QAP

$\text{N}\text{p}\to \text{P}{\text{a+}}_2^4\alpha$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the production of Pa-233 through alpha emission.

Pa-233 is the isotope of protactinium, atomic number of protactinium is 91.

Alpha emission is a radioactive decay in which the nucleus (unstable) emits an alpha particle

( ${}_2^4\alpha$) and thus, results in the formation of different nucleus having atomic number which is reduced by 2 and mass number which is reduced by 4.

To identify the missing species: Let the species be $\text{X}$

The reaction can be written as:

${}_Z^A\text{X}\to \text{P}{\text{a+}}_2^4\alpha$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}A=233+4\\ A=237\end{array}$

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}Z=91+2\\ Z=93\end{array}$

Thus, the reactant must have mass number and atomic number equal to 237 and 93. From the periodic table, the nucleus is $\text{N}\text{p}$.

Overall reaction is:

$\text{N}\text{p}\to \text{P}{\text{a+}}_2^4\alpha$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The positron is lost by Y-85.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 11QAP

$\text{Y}\to e\text{+}\text{S}\text{r}$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the positron is lost by Y-85.

Y-85 is the isotope of yttrium, atomic number of yttrium is 39.

A radioactive decay in which a proton presents inside a nucleus (radionuclide) is changed into a neutron during releasing an electron neutrino and a positron is known as positron emission. The resulting nucleus has atomic number which is reduced by one and mass number will remain same.

To identify the missing species: Let the species be $\text{X}$

The reaction can be written as:

$\text{Y}\to e{\text{+}}_Z^A\text{X}$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}85=0+A\\ A=85\end{array}$

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}39=1+Z\\ Z=38\end{array}$

Thus, the product must have mass number and atomic number equal to 85 and 38. From the periodic table, the nucleus is $\text{S}\text{r}$.

Overall reaction is:

$\text{Y}\to e\text{+}\text{S}\text{r}$

Interpretation Introduction

(c)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The production of Na-23 and another particle from the fusion two nuclei of C-12.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 11QAP

$2\text{C}\to \text{N}\text{a+}\text{H}$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the production of Na-23 and another particle from the fusion two nuclei of C-12.

Na-23 is the isotope of sodium, atomic number of sodium is 11.

C-12 is the isotope of carbon, atomic number of carbon is 6.

To identify the missing species: Let the species be $\text{X}$

The reaction can be written as:

$2\text{C}\to \text{N}\text{a+}\text{X}$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}24=23+A\\ A=1\end{array}$, thus mass number on both sides are equal

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}12=11+Z\\ Z=1\end{array}$, thus charges are equal on both sides

Thus, another particle belongs to nuclei of hydrogen atom that is proton.

Overall reaction is:

$2\text{C}\to \text{N}\text{a+}\text{H}$

Interpretation Introduction

(d)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The production of Sn-130 with another nucleus and excess of two neutrons from the fusion of

Pu-239.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 11QAP

${}_{94}^{239}\text{Pu}\to \text{S}\text{n+}\text{R}{\text{u+2}}_0^{1}n$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the production of Sn-130 with another nucleus and excess of two neutrons from the fusion of Pu-239.

Sn-130 is the isotope of tin, atomic number of tin is 50.

Pu-239 is the isotope of plutonium, atomic number of plutonium is 94.

Now, symbol of neutron is ${}_0^{1}n$.

To identify the missing species: Let the species be $\text{X}$

The reaction can be written as:

${}_{94}^{239}\text{Pu}\to \text{S}\text{n+}\text{X}{\text{+2}}_0^{1}n$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}239=130+A+2\\ A=239-132=107\end{array}$

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}94=50+Z+0\\ Z=94-50=44\end{array}$

Thus, the product must have mass number and atomic number equal to 107 and 44. From the periodic table, the nucleus is $\text{R}\text{u}$.

Overall reaction is:

${}_{94}^{239}\text{Pu}\to \text{S}\text{n+}\text{R}{\text{u+2}}_0^{1}n$