Chapter 25, Problem 44A

(a)

Interpretation Introduction

Interpretation:

More stable isotopes are to be identified between $\text{C}$ & $\text{C}$

Concept introduction:

The three primary isotopes of carbon are carbon-12, carbon-13, and carbon-14.

Answer to Problem 44A

  $\text{C}$ is a more stable isotope.

Explanation of Solution

The stability of the nucleus and isotope depends on the ratio of neutrons to protons. When we compare two different isotopes, a nucleus with a neutron-to-proton ratio closer to 1 is more stable because this isotope lies closer to the band of stability.

• $\text{C}$ Isotope has $\text{6}$ protons (it is the atomic number) and $\text{8}$ neutrons $\text{(14-6=8)}$ and the neutron-to-proton ratio is:

  $\text{ratio=}\frac{{\text{n(n}}^{\text{0}}\text{)}}{{\text{n(p}}^{\text{+}}\text{)}}\text{=}\frac{\text{8}}{\text{6}}\text{=1}\text{.33}$

• $\text{C}$ Isotope has $\text{6}$ protons (it is the atomic number) and $7$ neutrons $\text{(13-6=7)}$ and the neutron-to-proton ratio is:

  $\text{ratio=}\frac{{\text{n(n}}^{\text{0}}\text{)}}{{\text{n(p}}^{\text{+}}\text{)}}\text{=}\frac{7}{\text{6}}\text{=1}\text{.17}$

Hence,

  $\text{C}$ has a neutron-to-proton ratio of $\text{1}\text{.33}$ while $\text{C}$ has a neutron-to-proton ratio of $\text{1}\text{.17}$ . Therefore $\text{C}$ is more stable, because a neutron-to-proton ratio closer to $\text{1}$ lies closer to the band of stability.

(b)

Interpretation Introduction

Interpretation:

More stable isotope between $\text{H}\text{, and}\text{H}$ is to be identified.

Concept Interpretation:

Protium $\text{H}$ , deuterium $\text{H}$ or D, and tritium $\text{H}$ or T are the three hydrogen isotopes.

Answer to Problem 44A

  $\text{H}$ is a more stable isotope.

Explanation of Solution

The stability of the nucleus and isotope depends on the ratio of neutrons to protons. When we compare two different isotopes, a nucleus with a neutron-to-proton ratio closer to 1 is more stable because this isotope lies closer to the band of stability.

• $H$ Isotope has $1$ protons (it is the atomic number) and $2$ neutrons $\text{(3-1=2)}$ and the neutron-to-proton ratio is:

  $\text{ratio=}\frac{{\text{n(n}}^{\text{0}}\text{)}}{{\text{n(p}}^{\text{+}}\text{)}}\text{=}\frac{2}{1}\text{=2}\text{.0}$

• $H$ Isotope has $1$ protons (it is the atomic number) and $0$ neutrons $\text{(1-1=0)}$ and the neutron-to-proton ratio is:

  $\text{ratio=}\frac{{\text{n(n}}^{\text{0}}\text{)}}{{\text{n(p}}^{\text{+}}\text{)}}\text{=}\frac{0}{1}\text{=0}$

Hence,

  $\text{H}$ has a neutron-to-proton ratio of $\text{2}$ while $\text{H}$ has a neutron-to-proton ratio of $0$ . Therefore, $\text{H}$ is more stable, because a neutron-to-proton ratio of $0$ lies in the band of stability.

(c)

Interpretation Introduction

Interpretation:

More stable isotope between $\text{O}\text{,}\text{O}$ is to be identified.

Concept Interpretation:

The oxygen element $\left(\text{O}\right)$ has three isotopes: oxygen $16$ , oxygen $17$ , and oxygen $\text{18}$

Answer to Problem 44A

  $\text{O}$ is a more stable isotope.

Explanation of Solution

The stability of the nucleus and isotope depends on the ratio of neutrons to protons. When we compare two different isotopes, a nucleus with a neutron-to-proton ratio closer to 1 is more stable because this isotope lies closer to the band of stability.

• $O$ Isotope has $8$ protons (it is the atomic number) and $7$ neutrons $\text{(15-8=7)}$ and the neutron-to-proton ratio is:

  $\text{ratio=}\frac{{\text{n(n}}^{\text{0}}\text{)}}{{\text{n(p}}^{\text{+}}\text{)}}\text{=}\frac{7}{8}\text{=0}\text{.88}$

• $O$ Isotope has $8$ protons (it is the atomic number) and $8$ neutrons $\text{(16-8=8)}$ and the neutron-to-proton ratio is:

  $\text{ratio=}\frac{{\text{n(n}}^{\text{0}}\text{)}}{{\text{n(p}}^{\text{+}}\text{)}}\text{=}\frac{8}{8}\text{=1}\text{.00}$

Hence,

  $\text{O}$ has a neutron-to-proton ratio of 1 while $\text{O}$ has a neutron-to-proton ratio of $0.88$ . Therefore $\text{O}$ is more stable, because a neutron-to-proton ratio closer $\text{1}$ lies closer to the band of stability.

(d)

Interpretation Introduction

Interpretation:

More stable isotope between $\text{N}\text{,}\text{N}$ is to be identified.

Concept Interpretation:

There are three isotopes of nitrogen such that:- nitrogen- $13$ , nitrogen- $14$ , and nitrogen- $15$ ,

Answer to Problem 44A

  $\text{N}$ is more stable.

Explanation of Solution

The stability of the nucleus and isotope depends on the ratio of neutrons to protons. When we compare two different isotopes, a nucleus with a neutron-to-proton ratio closer to 1 is more stable because this isotope lies closer to the band of stability.

• $\text{N}$ isotope has a mass number of $\text{14}$ and an atomic number of $\text{7}$ , so, the number of protons and neutrons is:

  $\begin{array}{c}\text{n}\left({\text{p}}^{\text{+}}\right)\text{=A=7}\\ \text{n}\left({\text{n}}^{\text{0}}\right)\text{=A-Z=15-7=8}\end{array}$

The ratio of neutrons to protons in $\text{N}$ is:

  $\text{ratio=}\frac{\text{n}\left({\text{n}}^{\text{0}}\right)}{\text{n}\left({\text{p}}^{\text{+}}\right)}\text{=}\frac{\text{7}}{\text{7}}\text{=1}$

• $\text{N}$ isotope has a mass number of $\text{15}$ and an atomic number of $\text{7}$ , so, the number of protons and neutrons is:

  $\begin{array}{c}\text{n}\left({\text{p}}^{\text{+}}\right)\text{=A=7}\\ \text{n}\left({\text{n}}^{\text{0}}\right)\text{=A-Z=15-7=8}\end{array}$

The ratio of neutrons to protons in $\text{N}$ is :

  $\text{ratio=}\frac{\text{n}\left({\text{n}}^{\text{0}}\right)}{\text{n}\left({\text{p}}^{\text{+}}\right)}\text{=}\frac{8}{\text{7}}\text{=1}\text{.14}$

Hence, $\text{N}$ isotope has a ratio of neutrons and protons closer to $\text{1}$ , it is more stable.