Chapter 18, Problem 17QAP

Interpretation Introduction

(a)

Interpretation:

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

The bombardment of U-238 with a nucleus results in the production of Fm-249 with five neutrons.

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 17QAP

$\text{U}\text{ + }\text{O}{\to }_{\text{100}}^{\text{249}}{\text{Fm+5}}_0^{1}n$

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 bombardment of U-238 with a nucleus results in the production of Fm-249 with five neutrons.

U-238 is the isotope of uranium, atomic number of uranium is 92.

Fm-249 is the isotope of fermium, atomic number of fermium is 100.

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

The reaction can be written as:

$\text{U}{\text{ + }}_Z^A\text{X}{\to }_{\text{100}}^{\text{249}}{\text{Fm+5}}_0^{1}n$

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

$\begin{array}{l}238+A=249+5\\ A=254-238=16\end{array}$

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

$\begin{array}{l}92+Z=100\\ Z=8\end{array}$

Thus, the product must have mass number and atomic number equal to 16 and 8. From the periodic table, the nucleus is $\text{O}$.

Overall reaction is:

$\text{U}\text{ + }\text{O}{\to }_{\text{100}}^{\text{249}}{\text{Fm+5}}_0^{1}n$

Interpretation Introduction

(b)

Interpretation:

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

The bombardment of Al-26 with an alpha particle results in the production of P-30.

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 17QAP

${}_{13}^{\text{26}}{\text{Al+}}_2^4\alpha \to P$

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 bombardment of Al-26 with an alpha particle results in the production of P-30.

Al-26 is the isotope of aluminium, atomic number of aluminium is 13.

P-30 is the isotope of phosphorus, atomic number of phosphorus is 15.

The reaction can be written as:

${}_{13}^{\text{26}}{\text{Al+}}_2^4\alpha \to P$

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

$\begin{array}{l}26+4=A\\ A=30\end{array}$, thus, mass number is balanced

Now, atomic number or charges must be balanced, let Z be the atomic number of phosphorus, therefore:

$\begin{array}{l}13+2=Z\\ Z=15\end{array}$, thus atomic number is balanced

Overall reaction is:

${}_{13}^{\text{26}}{\text{Al+}}_2^4\alpha \to P$

Interpretation Introduction

(c)

Interpretation:

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

The bombardment of Cu-63 with a nucleus results in the production of Zn-63 with a neutron.

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 17QAP

${}_{29}^{63}\text{Cu +}\text{H}\to \text{Z}{\text{n+}}_0^{1}n$

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 bombardment of Cu-63 with a nucleus results in the production of Zn-63 with a neutron.

Cu-63 is the isotope of copper, atomic number of copper is 29.

Zn-63 is the isotope of zinc, atomic number of zinc is 30.

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:

${}_{29}^{63}\text{Cu + }\text{X}\to \text{Z}{\text{n+}}_0^{1}n$

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

$\begin{array}{l}63+A=63+1\\ A=64-63=1\end{array}$

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

$\begin{array}{l}29+Z=30+0\\ Z=30-29=1\end{array}$

Thus, the reactant must have mass number and atomic number equal to 1 and 1. From the periodic table, the nucleus is $\text{H}$.

Overall reaction is:

${}_{29}^{63}\text{Cu +}\text{H}\to \text{Z}{\text{n+}}_0^{1}n$

Interpretation Introduction

(d)

Interpretation:

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

The bombardment of Al-27 with ${}_1^2\text{H}$ results in the production of an alpha particle with another nucleus.

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 17QAP

${}_{13}^{27}{\text{Al+}}_1^2\text{H}\to \text{M}{\text{g+}}_2^4\alpha$

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 bombardment of Al-27 with ${}_1^2\text{H}$ results in the production of an alpha particle with another nucleus.

Al-27 is the isotope of aluminium, atomic number of aluminium is 13.

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

The reaction can be written as:

${}_{13}^{27}{\text{Al+}}_1^2\text{H}{\to }_Z^A{\text{X+}}_2^4\alpha$

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

$\begin{array}{l}27+2=A+4\\ A=29-4=25\end{array}$

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

$\begin{array}{l}13+1=Z+2\\ Z=14-2=12\end{array}$

Thus, the product must have mass number and atomic number equal to 25 and 12. From the periodic table, the nucleus is $\text{M}\text{g}$.

Overall reaction is:

${}_{13}^{27}{\text{Al+}}_1^2\text{H}\to \text{M}{\text{g+}}_2^4\alpha$