Chapter 45, Problem 48SP

Complete the notations for the following processes.

(a) ${}^{24}\text{Mg}\left(d,p\right)?$

(b) ${}^{26}\text{Mg}\left(d,p\right)$?

(c) ${}^{40}\text{Ar}\left(\alpha ,p\right)$?

(d) ${}^{12}\text{C}\left(d,n\right)$?

(e) ${}^{130}\text{Te}\left(d,2n\right)$?

(f) ${}^{55}\text{Mn}\left(n,\gamma \right)$?

(g) ${}^{59}\text{Co}\left(n,\alpha \right)$?

To determine

To complete: The notation of the decay process for magnesium, if the alpha particles are emitted in the decay process as, ${}^{24}\text{Mg}\left(d,\alpha \right)$.

Answer to Problem 48SP

Solution:

${}^{24}\text{Mg}\left(d,\alpha \right)\text{N}\text{a}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$

Explanation:

Write the decay equation for ${}^{24}\text{Mg}\left(d,\alpha \right)\text{X}$ to complete the notation of the decay. Here particle $d$ is the deuteron $\left({}_1^2\text{H}\right)$, and $\alpha$ stands for the alpha particle.

${}_{12}^{24}{\text{Mg+}}_1^2\text{H}\stackrel{}{\to }{}_b^a{\text{X+}}_2^4\alpha$

Here, $\text{X}$ is the element of the atomic mass number $a$ and the number of protons is $b$.

The sum of the mass number of the reactants and products are equal. Hence,

$24+2=a+4$

Solve for $a$.

$a=22$

The sum of the atomic number of the reactants and products are equal. Hence,

$12+1=b+2$

Solve for $b$.

$b=11$

The sodium element has atomic number $11$ and mass number $22$ .

Thus, rewrite the notation.

${}^{24}\text{Mg}\left(d,\alpha \right)\text{X}$

Substitute $\text{Na}$ for $\text{X}$ and $22$ for $a$

${}^{24}\text{Mg}\left(d,\alpha \right)\text{N}\text{a}$.

Conclusion:

Therefore, the notations of the process are ${}^{24}\text{Mg}\left(d,\alpha \right)\text{N}\text{a}$.

To determine

To complete: The notation of the decay process for magnesium of the mass number $26$, if the proton is emitted in the decay process.

Answer to Problem 48SP

Solution:

${}^{26}\text{Mg}\left(d,\alpha \right)\text{M}\text{g}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$

Explanation:

Write the decay equation for ${}^{26}\text{Mg}\left(d,p\right)\text{X}$ to complete the notation of the decay. Here particle $d$ is the deuteron $\left({}_1^2\text{H}\right)$, and $p$ stands for the proton.

${}_{12}^{26}{\text{Mg+}}_1^2\text{H}\stackrel{}{\to }{}_b^a{\text{X+}}_1^1\text{H}$

Here, $\text{X}$ is the element of the atomic mass number $a$ and thenumber of protons is $b$.

The sum of the mass number of the reactants and products are equal. Hence,

$26+2=a+1$

Solve for $a$.

$a=27$

The sum of the atomic number of the reactants and products are equal. Hence,

$12+1=b+1$

Solve for $b$.

$b=12$

The magnesium element has an atomic number thatequals to $12$.

Rewrite the notation.

${}^{26}\text{Mg}\left(d,p\right)\text{X}$

Substitute $\text{Mg}$ for $\text{X}$ and $27$ for $a$

${}^{26}\text{Mg}\left(d,\alpha \right)\text{M}\text{g}$.

Conclusion:

Therefore, the notations of the process are ${}^{26}\text{Mg}\left(d,\alpha \right)\text{M}\text{g}$.

To determine

To complete: The notation of the decay process for argon of mass number equals to $40$ given as ${}^{40}\text{Ar}\left(\alpha ,p\right)$.

Answer to Problem 48SP

Solution:

${}^{40}\text{Ar}\left(\alpha ,p\right)\text{K}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$

Explanation:

Write the decay equation for ${}^{40}\text{Ar}\left(\alpha ,p\right)\text{X}$ to complete the notation of the decay. Here, particle $\alpha$ stands for the alpha particle and $p$ stands for the proton.

${}_{18}^{40}{\text{Ar+}}_2^4\alpha \stackrel{}{\to }{}_b^a{\text{X+}}_1^1\text{H}$

Here, $\text{X}$ is the element of the atomic mass number $a$ and the number of protons is $b$.

The sum of the mass number of the reactants and products are equal. Hence,

$40+4=a+1$

Solve for $a$.

$a=43$

The sum of the atomic number of the reactants and products are equal. Hence,

$18+2=b+1$

Solve for $b$.

$b=19$

The potassium element has atomic number $19$ and mass number $43$ .

Rewrite the notation.

${}^{40}\text{Ar}\left(\alpha ,p\right)\text{X}$

Substitute $\text{K}$ for $\text{X}$ and $43$ for $a$

${}^{40}\text{Ar}\left(\alpha ,p\right)\text{K}$

Conclusion:

Therefore, the notations of the process are ${}^{40}\text{Ar}\left(\alpha ,p\right)\text{K}$.

To determine

To complete: The notation of the decay process forcarbon of the mass $12$ given as ${}^{12}\text{C}\left(d,n\right)$, if the neutron is ejected.

Answer to Problem 48SP

Solution:

${}^{12}\text{C}\left(d,n\right)\text{N}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$

Explanation:

Write the decay equation for ${}^{12}\text{C}\left(d,n\right)\text{X}$ to complete the notation of the decay. Here, particle $d$ stands for the deuteron $\left({}_1^2\text{H}\right)$ and $n$ stands for the neutron.

${}_6^{12}{\text{C +}}_1^2\text{H}\stackrel{}{\to }{}_b^a{\text{X+}}_0^{1}n$

Here, $\text{X}$ is the element of the atomic mass number $a$ and the number of protons is $b$.

The sum of the mass number of the reactants and products are equal. Hence,

$12+2=a+1$

Solve for $a$.

$a=13$

The sum of the atomic number of the reactants and products are equal. Hence,

$6+1=b+0$

Solve for $b$.

$b=7$

The nitrogen element has atomic number $7$ and mass number $13$ .

Rewrite the notation.

${}^{12}\text{C}\left(d,n\right)\text{X}$

Substitute $\text{N}$ for $\text{X}$ and $13$ for $a$

${}^{12}\text{C}\left(d,n\right)\text{N}$

Conclusion:

Therefore, the notations of the process are ${}^{12}\text{C}\left(d,n\right)\text{N}$.

To determine

To complete: The notation of the decay process for tellurium $\text{Te}$ of mass number $130$ given as ${}^{130}\text{Te}\left(d,2n\right)$, if two neutrons are ejected.

Answer to Problem 48SP

Solution:

${}^{130}\text{Te}\left(d,2n\right)\text{I}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$

Explanation:

Write the decay equation for ${}^{24}\text{Te}\left(d,2n\right)\text{X}$ to complete the notation of the decay. Here, particle $d$ stands for the deuteron $\left({}_1^2\text{H}\right)$, and $n$ stands for the neutron.

${}_{52}^{130}{\text{Te+}}_1^2\text{H}\stackrel{}{\to }{}_b^a{\text{X+2}}_0^{1}n$

The sum of the mass number of the reactants and products are equal. Hence,

$130+2=a+2$

Solve for $a$.

$a=130$

The sum of the atomic number of the reactants and products are equal. Hence,

$52+1=b+2\left(0\right)$

Solve for $b$.

$b=53$

The iodine element has atomic number $53$ and mass number $130$ .

Rewrite the notation.

${}^{130}{\text{T}}_e\left(d,2n\right)\text{X}$

Substitute $\text{I}$ for $\text{X}$ and $130$ for $a$

${}^{130}\text{Te}\left(d,2n\right)\text{I}$

Conclusion:

Therefore, the notations of the process are ${}^{130}\text{Te}\left(d,2n\right)\text{I}$.

To determine

To complete: The notation of the decay process formanganese of mass number $55$ given by, if gamma particle is separated in the decay process.

Answer to Problem 48SP

Solution:

${}^{55}\text{Mn}\left(n,\gamma \right)\text{M}\text{n}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$.

Explanation:

Write the decay equation for ${}^{55}\text{Mn}\left(n,\gamma \right)\text{X}$ to complete the notation of the decay. Here, particle $n$ stands for the neutron, and $\gamma$ stands for the gamma particle.

${}_{25}^{55}{\text{Mn+}}_0^1\text{n}\stackrel{}{\to }{}_b^a\text{X+}\gamma$

The sum of the mass number of the reactants and products are equal. Hence,

$55+1=a$

Solve for $a$.

$a=56$

The sum of the atomic number of the reactants and products are equal. Hence,

$\begin{array}{l}25=b\\ b=25\end{array}$

The manganese element has atomic number $25$ and mass number $55$ .

Rewrite the notation.

${}^{55}\text{Mn}\left(n,\gamma \right)\text{X}$

Substitute $\text{Mn}$ for $\text{X}$ and $56$ for $a$

${}^{55}\text{Mn}\left(n,\gamma \right)\text{M}\text{n}$

Conclusion:

Therefore, the notations of the process are ${}^{55}\text{Mn}\left(n,\gamma \right)\text{M}\text{n}$.

To determine

To complete: The notation of the decay process forcobalt of mass number $59$ given by ${}^{59}\text{Co}\left(n,\alpha \right)$, if alpha particle is separated in the decay process.

Answer to Problem 48SP

Solution:

${}^{59}\text{Co}\left(n,\alpha \right)\text{M}\text{n}$

Explanation of Solution

Introduction:

The decay process equation is written as,

${}_b^a\text{A}$+$\text{D}$ $\underset{}{\to }$ ${}_d^c\text{B}$+$\text{E}$

The notation for decay process is written as ${}_b^a\text{A}{\left(D,E\right)}_f^c\text{B}$, where $\text{D}$ and $\text{E}$ are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, $a+e=c+g$ and $b+f=d+h$.

Explanation:

Write the decay equation for ${}^{59}\text{Co}\left(n,\alpha \right)\text{X}$ to complete the notation of the decay. Here, particle $n$ stands for the neutron, and $\alpha$ stands for the alpha particle.

${}_{27}^{59}{\text{Co+}}_0^1\text{n }\stackrel{}{\to }{}_b^a{\text{X+ }}_2^4\alpha$

The sum of the mass number of the reactants and products are equal. Hence,

$59+1=a+4$

Solve for $a$.

$a=56$

The sum of the atomic number of the reactants and products are equal. Hence,

$27+0=b+2$

Solve for $b$.

$b=25$

The manganese element has atomic number $25$ and mass number $56$ .

Rewrite the notation.

${}^{59}\text{Co}\left(n,\alpha \right)\text{X}$

Substitute $\text{Mn}$ for $\text{X}$ and $56$ for $a$

${}^{59}\text{Co}\left(n,\alpha \right)\text{M}\text{n}$

Conclusion:

Therefore, the notations of the process are ${}^{59}\text{Co}\left(n,\alpha \right)\text{M}\text{n}$.