Chapter 45, Problem 30P

(a)

To determine

Identify nucleus $A$.

Answer to Problem 30P

The nucleus $A$ is nitrogen and its symbol is ${}_7^{13}\text{N}$.

Explanation of Solution

A highly energized proton is absorbed by the ${}_6^{12}\text{C}$ and produces an unknown nucleus $A$ and a gamma ray.

    ${}_1^1\text{H}{+}_6^{12}\text{C}{\to }_{?}^{?}\text{A}+\gamma$

The atomic number of  ${}_1^1\text{H}$ is $Z=1$ and its mass number is $A=1$, and the atomic number of ${}_6^{12}\text{C}$ is $Z=6$ and its mass number $A=12$. Thus, the atomic and mass number for the unknown nucleus $A$ is $Z=1+6=7$ and $A=1+12=13$ respectively.

In periodic table, the element with atomic number $Z=7$ is nitrogen, ${}_7^{13}\text{N}$. The reaction becomes,

    ${}_1^1\text{H}{+}_6^{12}\text{C}{\to }_7^{13}\text{N}+\gamma$

Thus, the nucleus $A$ is nitrogen and its symbol is ${}_7^{13}\text{N}$.

(b)

To determine

Identify nucleus $B$.

Answer to Problem 30P

The nucleus $B$ is carbon and its symbol is ${}_6^{13}\text{C}$.

Explanation of Solution

As ${}_7^{13}\text{N}$ is the isotope of nitrogen, it decays by emitting a positron and a neutrino.

    ${}_7^{13}\text{N}\to ?{+}_1^0\text{e}+\nu$

The atomic and mass number for the nitrogen is $Z=7$ and $A=13$ respectively. Thus, the atomic and mass number of unknown nucleus $B$ is $Z=7-1=6$ and $A=13-0=13$.

In periodic table, the element with atomic number $Z=6$ is carbon, ${}_6^{13}\text{C}$. The reaction becomes,

    ${}_7^{13}\text{N}{\to }_6^{13}\text{C}{+}_1^0\text{e}+\nu$

Thus, the nucleus $B$ is carbon and its symbol is ${}_6^{13}\text{C}$.

(c)

To determine

Identify nucleus $C$.

Answer to Problem 30P

The nucleus $C$ is nitrogen and its symbol is ${}_7^{14}\text{N}$.

Explanation of Solution

${}_6^{13}\text{C}$ nucleus absorbs a proton and produce nucleus $C$ along with the emission of gamma ray.

    ${}_1^1\text{H}{+}_6^{13}\text{C}\to ?+\gamma$

The atomic number of  ${}_1^1\text{H}$ is $Z=1$ and its mass number is $A=1$, and the atomic number of ${}_6^{13}\text{C}$ is $Z=6$ and its mass number $A=13$. Thus, the atomic and mass number for the unknown nucleus $A$ is $Z=1+6=7$ and $A=1+13=14$ respectively.

In periodic table, the element with atomic number $Z=7$ is nitrogen, ${}_7^{14}\text{N}$. The reaction becomes,

    ${}_1^1\text{H}{+}_6^{13}\text{C}{\to }_7^{14}\text{N}+\gamma$

Thus, the nucleus $C$ is nitrogen and its symbol is ${}_7^{14}\text{N}$.

(d)

To determine

Identify nucleus $D$ .

Answer to Problem 30P

The nucleus $D$ is oxygen and its symbol is ${}_8^{15}\text{O}$.

Explanation of Solution

A proton is absorbed by the ${}_7^{14}\text{N}$ and produces an unknown nucleus $D$ and a gamma ray.

    ${}_1^1\text{H}{+}_7^{14}\text{N}\to ?+\gamma$

The atomic number of  ${}_1^1\text{H}$ is $Z=1$ and its mass number is $A=1$, and the atomic and mass number for the nitrogen is $Z=7$ and $A=14$ respectively. Thus, the atomic and mass number of unknown nucleus $D$ is $Z=7+1=8$ and $A=14+1=15$.

In periodic table, the element with atomic number $Z=8$ is oxygen, ${}_8^{15}\text{O}$. The reaction becomes,

    ${}_1^1\text{H}{+}_7^{14}\text{N}{\to }_8^{15}\text{O}+\gamma$

Thus, the nucleus $D$ is oxygen and its symbol is ${}_8^{15}\text{O}$.

(e)

To determine

Identify nucleus $E$.

Answer to Problem 30P

The nucleus $E$ is nitrogen and its symbol is ${}_7^{15}\text{N}$.

Explanation of Solution

As ${}_8^{15}\text{O}$ is the isotope of oxygen, it decays by emitting a positron and a neutrino.

    ${}_8^{15}\text{O}\to ?{+}_1^0\text{e}+\nu$

The atomic and mass number for the nitrogen is $Z=8$ and $A=18$ respectively. Thus, the atomic and mass number of unknown nucleus $B$ is $Z=8-1=7$ and $A=15-0=15$.

In periodic table, the element with atomic number $Z=7$ is nitrogen, ${}_7^{15}\text{N}$. The reaction becomes,

    ${}_8^{15}\text{O}{\to }_7^{15}\text{N}{+}_1^0\text{e}+\nu$

Thus, the nucleus $E$ is nitrogen and its symbol is ${}_7^{15}\text{N}$.

(f)

To determine

Identify nucleus $F$.

Answer to Problem 30P

The nucleus $F$ is carbon and its symbol is ${}_6^{12}\text{C}$.

Explanation of Solution

A ${}_7^{15}\text{N}$ nucleus absorbs a proton and produces an unknown nucleus $F$ and alpha particle.

    ${}_1^1\text{H}{+}_7^{15}\text{N}\to ?{+}_2^4\text{He}$

The atomic number of  ${}_1^1\text{H}$ is $Z=1$ and its mass number is $A=1$, and the atomic and mass number for the nitrogen is $Z=7$ and $A=15$ respectively. Thus, the atomic and mass number of unknown nucleus $F$ is $Z=7+1-2=6$ and $A=15+1-4=12$.

In periodic table, the element with atomic number $Z=6$ is carbon, ${}_6^{12}\text{C}$. The reaction becomes,

    ${}_1^1\text{H}{+}_7^{15}\text{N}{\to }_6^{12}\text{C}{+}_2^4\text{He}$

Thus, the nucleus $F$ is carbon and its symbol is ${}_6^{12}\text{C}$.

(g)

To determine

The significance of the final nucleus in the last step of the carbon cycle.

Answer to Problem 30P

The significance of the final nucleus in the last step of the carbon cycle is that ${}_6^{12}\text{C}$ is used as a catalyst.

Explanation of Solution

The steps of the carbon cycle reactions in part (a)-(f) are

1. (a) ${}_1^1\text{H}{+}_6^{12}\text{C}{\to }_7^{13}\text{N}+\gamma$
2. (b) ${}_7^{13}\text{N}{\to }_6^{13}\text{C}{+}_1^0\text{e}+\nu$
3. (c) ${}_1^1\text{H}{+}_6^{13}\text{C}{\to }_7^{14}\text{N}+\gamma$
4. (d) ${}_1^1\text{H}{+}_7^{14}\text{N}{\to }_8^{15}\text{O}+\gamma$
5. (e) ${}_8^{15}\text{O}{\to }_7^{15}\text{N}{+}_1^0\text{e}+\nu$
6. (f) ${}_1^1\text{H}{+}_7^{15}\text{N}{\to }_6^{12}\text{C}{+}_2^4\text{He}$

The overall reaction is ${}_1^1\text{H}{+}_6^{12}{\text{C+}}_1^1\text{H}{+}_1^1\text{H}{+}_1^1\text{H}+2_{-1}^{0}e{\to }_6^{12}\text{C}{+}_2^4\text{He}+7\gamma +\nu$. The positrons emitted in these reactions are annihilated immediately by free electrons, ${}_{-1}^{0}e{+}_{+1}^{0}e\to 2\gamma$.

Simplifying the equation, $4_1^1\text{H}+2_{-1}^{0}e{\to }_2^4\text{He}+\nu$. In the overall reaction, the nucleus $F$ in part (f), ${}_6^{12}\text{C}$ is used as a catalyst in the carbon cycle reaction.