Q. 5.63 - An Introduction to General, Organic, and Biological Chemistry

Question

Complete each of the following nuclear equations: $(5.2)$

a. ${}_2{}^4\mathrm{He}+{}_7{}^{14} \mathrm{N}⟶?+{}_1{}^1\mathrm{H}$

b. ${}_2{}^4\mathrm{He}+{}_{13}{}^{27}\mathrm{Al}⟶{}_{14}{}^{30}\mathrm{Si}+?$

c. ${}_0{}^{1}n+{}_{92}{}^{235}\mathrm{U}⟶{}_{38}{}^{90}\mathrm{Sr}+3{}_0{}^{1}n+?$

d. ${}_{12}{}^{23 \mathrm{m}}\mathrm{Mg}⟶$ $?+{}_0{}^0\gamma$

Step-by-Step Solution

Answer

The nuclear reactions are

(part a)  ${}_2{}^4\mathrm{He}+{}_7{}^{14} \mathrm{N}\to {}_8{}^{17}\mathrm{O}+{}_1{}^1\mathrm{H}$

(part b)   ${}_2{}^4\mathrm{He}+{}_{13}{}^{27}\mathrm{Al}\to {}_{14}{}^{30}\mathrm{Si}+{}_1{}^1\mathrm{H}.$

(part c) ${}_0{}^{1}n+{}_{92}{}^{235}\mathrm{U}\to {}_{38}{}^{{}^{90}}\mathrm{Sr}+3_0^{1}n+{}_4{}^{143}\mathrm{Xe}$

(part d) ${}_{12}{}^{23 \mathrm{m}}\mathrm{Mg}\to {}_{12}{}^{23}\mathrm{Mg}+{}_0{}^0\gamma$

Step 1- Given information (part a)

(part a) 

Alpha particles collide with nitrogen atoms. Alpha particles are the same as helium nuclei.  

Incomplete core equations such as:

${}_2{}^4\mathrm{He}+{}_7{}^{14} \mathrm{N}\to ?+{}_1{}^1\mathrm{H}$

Find the missing mass number in the above equation. 

$\text{ Mass number of new nucleus }=\text{ (mass number of Helium }+\text{ mass number of nitrogen })$

                                                         $- (mass number of hydrogen atom)$

Substitute the value of the above expression. 

$Mass number of new nucleus =(4+14)-\left(1\right)$

                                                            $=17$

Step 2- Given information (part b)

Find the missing atomic  number in the above equation. 

$\text{ Atomic number of new nucleus = (atomic number of Helium }+\text{ atomic number of nitrogen) }$

                                                            $- (atomic number of hydrogen atom)$

Substitute the value of the above expression. 

$Atomic number of new nucleus =(2+7)-\left(1\right)$

                                                             $=8$

Determine the symbol for the new element.

In the  periodic table , atomic number 8 represents oxygen $\left(O\right)$

The isotope symbol is written as ${}_3{}^{17}\mathrm{O}$

The complete nuclear reaction is: 

${}_2{}^4\mathrm{He}+{}_7{}^{14} \mathrm{N}\to {}_8{}^{17}\mathrm{O}+{}_1{}^1\mathrm{H}$

. 

Step 3- Explanation-1 (Part a)

(part a)

Incomplete core equations such as

${}_2{}^4\mathrm{He}+{}_{13}{}^{27}\mathrm{Al}\to {}_{14}{}^{30}\mathrm{Si}+?$

Find the missing mass number in the above equation 

$\text{ Mass number of new nucleus }=\text{ ( mass number of Helium }+\text{ mass number of aluminium })$

                                                     $- (mass number of silicon atom)$

Substitute the value of the above expression. 

$Mass number of new nucleus =(4+27)-\left(30\right)$ $=1$

Step 4- Explanation -1 (part b)

(part b) Find the missing atomic  number in the above equation. 

$\text{ Atomic number of new nucleus }=\left(\begin{array}{l}\text{ atomic number of Helium }\\ +\text{ atomic number of aluminium }\end{array}\right)$

                                                         $- (atomic number of silicon atom)$

Substitute the value of the above expression. 

$Atomic number of new nucleus =(2+13)-\left(14\right)$

                                                             $=1$.

Determine the symbol for the new element.

In the periodic table, atomic number 1 represents hydrogen $\left(\mathrm{H}\right)$

The isotope symbol is written as ${}_1{}^1\mathrm{H}$

The complete nuclear reaction is 

${}_2{}^4\mathrm{He}+{}_{13}{}^{27}\mathrm{Al}\to {}_{14}{}^{30}\mathrm{Si}+{}_1{}^1\mathrm{H}.$

Step 6- Explanation -2 (part c)

(part c)

Incomplete core equations such as

${}_0{}^{1}n+{}_{42}{}^{23}\mathrm{U}\to {}_{38}{}^{90}\mathrm{Sr}+3_0^{1}n+?$

Find the missing mass number in the above equation 

$\text{ Mass number of new nucleus }=(\text{ mass number of neutron }+\text{ mass number of U) }$

                                                        $-(\text{ mass number of }\mathrm{Sr}+\text{ mass number of }3n)$

Substitute the value of the above expression. 

$Mass number of new nucleus =(1+235)-(90+3)$

                                                              $=143$

Step 6- Explanation 2 (part d)

(part d) Find the missing atomic  number in the above equation. 

$\text{ Atomic number of new nucleus = (atomic number of neutron }+\text{ atomic number of U) }$

                                                        $-(\text{ atomic number of }\mathrm{Sr}+\text{ atomic number of }3n)$

Substitute the value of the above expression. 

$Atomic number of new nucleus =(0+92)-\left(38\right)$

                                                                   $=54$.

Determine the symbol for the new element.

In the periodic table, atomic number 54represents xenon $\left(\mathrm{Xe}\right)$

The isotope symbol is written as ${}_{54}{}^{143}\mathrm{Xe}$

The complete nuclear reaction is 

${}_0{}^{1}n+{}_{92}{}^{235}\mathrm{U}\to {}_{35}{}^{90}\mathrm{Sr}+3_0^{1}n+{}_4{}^{143}\mathrm{Xe}$

Step 7- Explanation 3 (part a)

(part a)

Incomplete core equations such as

${}_{12}{}^{23 \mathrm{m}}\mathrm{Mg}\to ?+{}_0{}^0\gamma$

Find the missing mass number in the above equation 

$\text{ Mass number of new nucleus }=\left(\text{ mass number of }\mathrm{Mg}\right)-\left(\text{ mass number of }\gamma \right)$

Substitute the value of the above expression. 

$Mass number of new nucleus =\left(23\right)-\left(0\right)$

                                                             $=23$

Step 4- Explanation 3 (part b)

(part b) Find the missing atomic  number in the above equation. 

$\text{ Atomic number of new nucleus }=\left(\text{ atomic number of }\mathrm{Mg}\right)-\left(\text{ atomic number of }\gamma \right)$

Substitute the value of the above expression. 

                                                         $=12$.

Determine the symbol for the new element.

In the periodic table, atomic number 12 represents magnesium $\left(\mathrm{Mg}\right)$

The isotope symbol is written as ${}_{12}{}^{23}\mathrm{Mg}$

The complete nuclear reaction is 

${}_{12}{}^{23 \mathrm{m}}\mathrm{Mg}\to {}_{12}{}^{23}\mathrm{Mg}+{}_0{}^0\gamma$