Chapter 30, Problem 18P

(a)

To determine

The nucleus A.

Answer to Problem 18P

The nucleus A is ${}_7^{13}N$.

Explanation of Solution

The reaction is,

$\begin{array}{l}{}_1^{1}H{+}_6^{12}C{\to }_7^{13}A+\gamma \\ \Rightarrow \\ {}_1^{1}H{+}_6^{12}C{\to }_7^{13}N+\gamma \end{array}$

Thus, the nucleus A is ${}_7^{13}N$.

Conclusion:

The nucleus A is ${}_7^{13}N$.

(b)

To determine

The nucleus B.

Answer to Problem 18P

The nucleus B is ${}_6^{13}C$.

Explanation of Solution

The reaction is,

$\begin{array}{l}{}_7^{13}N{\to }_6^{13}B{+}_{+1}^{0}e+\nu \\ \Rightarrow \\ {}_7^{13}N{\to }_6^{13}C{+}_{+1}^{0}e+\nu \end{array}$

Thus, the nucleus B is ${}_6^{13}C$.

Conclusion:

The nucleus B is ${}_6^{13}C$.

(c)

To determine

The nucleus C.

Answer to Problem 18P

The nucleus C is ${}_7^{14}N$.

Explanation of Solution

The reaction is,

$\begin{array}{l}{}_6^{13}C{+}_1^{1}H{\to }_7^{14}C+\gamma \\ \Rightarrow \\ {}_6^{13}C{+}_1^{1}H{\to }_7^{14}N+\gamma \end{array}$

Thus, the nucleus C is ${}_7^{14}N$.

Conclusion:

The nucleus C is ${}_7^{14}N$.

(d)

To determine

The nucleus D.

Answer to Problem 18P

The nucleus D is ${}_8^{15}O$.

Explanation of Solution

The reaction is,

$\begin{array}{l}{}_7^{14}N{+}_1^{1}H{\to }_8^{15}D+\gamma \\ \Rightarrow \\ {}_7^{14}N{+}_1^{1}H{\to }_8^{15}O+\gamma \end{array}$

Thus, the nucleus D is ${}_8^{15}O$.

Conclusion:

The nucleus D is ${}_8^{15}O$.

(e)

To determine

The nucleus E.

Answer to Problem 18P

The nucleus E is ${}_7^{15}N$.

Explanation of Solution

The reaction is,

$\begin{array}{l}{}_8^{15}O{+}_{-1}^{0}e{\to }_7^{15}E+\nu \\ \Rightarrow \\ {}_8^{15}O{+}_{-1}^{0}e{\to }_7^{15}N+\nu \end{array}$

Thus, the nucleus E is ${}_7^{15}N$.

Conclusion:

The nucleus E is ${}_7^{15}N$.

(f)

To determine

The nucleus F.

Answer to Problem 18P

The nucleus F is ${}_6^{12}C$.

Explanation of Solution

The reaction is,

$\begin{array}{l}{}_7^{15}N{+}_1^{1}H{\to }_6^{12}F{+}_2^{4}He\\ \Rightarrow \\ {}_7^{15}N{+}_1^{1}H{\to }_6^{12}C{+}_2^{4}He\end{array}$

Thus, the nucleus F is ${}_6^{12}C$.

Conclusion:

The nucleus F is ${}_6^{12}C$