Q. 5.64 - An Introduction to General, Organic, and Biological Chemistry

Question

Complete each of the following nuclear equations: 

1.  $?+\frac{59}{27}\mathrm{Co}⟶\frac{56}{25}\mathrm{Mn}+\frac{4}{2}\mathrm{He}$

       b. $?⟶{}_7{}^{14} \mathrm{N}+{}_{-1}{}^{0}e$

  

       c.${}_{-1}{}^{0}e+{}_{36}{}^{76}\mathrm{Kr}⟶?$

   

      d.${}_2{}^4\mathrm{He}+{}_{95}{}^{241}\mathrm{Am}⟶?+2{}_0{}^{1}n$

Step-by-Step Solution

Answer

The nuclear equations are

(part a)  ${}_0{}^{1}n+{}_{27}{}^{59}\mathrm{Co}\to {}_{25}{}^{56}\mathrm{Mn}+{}_2{}^4\mathrm{He}$

(part b)  ${}_6{}^{14}\mathrm{C}\to {}_7{}^{14} \mathrm{N}+{}_{-1}{}^{0}e$

(part c)  ${}_{-1}{}^{0}e+{}_{36}{}^{76}\mathrm{Kr}\to {}_{35}{}^{76}\mathrm{Br}$

(part d)   ${}_2{}^4\mathrm{He}+{}_{95}{}^{241}\mathrm{Am}\to {}_{97}{}^{247}\mathrm{Bk}+2_0^{1}n.$ 

Step 1- Given information (part a)

When an alpha particle  is emitted from an detail its nucleus loses  protons and  neutrons. 

When an alpha particle is emitted from the nucleus, 

the mass range decreases via way of means of four and atomic range decreases via way of means of $2$ .Alpha particle is same to helium nucleus.  

Incomplete nuclear equation is as follows:

$?+{}_{27}{}^{59}\mathrm{Co}\to {}_{25}{}^{56}\mathrm{Mn}+{}_2{}^4\mathrm{He}$

  

Step 2- Explanation (part a)

$\text{ Mass number of bombarded particle }=\left[\begin{array}{l}\text{ (mass number of Mn nucleus })\\ +\left(\text{ mass number of alpha particle }\right)\end{array}\right]$

$=-\left[\text{ mass number of }\mathrm{Co}\right]$

Substitute the value of the above expression. 

$Mass number of bombarded particle =\left[\right(56)+(4\left)\right]-\left[59\right]$ $=1$

Step 3- Explanation (part-a)

$\text{ Atomic number of bombarded particle }=\left[\begin{array}{l}\left(\text{ atomic number of Mn nucleus }\right)\\ +(\text{ atomic number of alpha particle) }\end{array}\right]$

  $=-[\text{ atomic number of Co]}$

Substitute the value of the above expression. 

$Atomic number of bombarded particle =\left[\right(25)+(2\left)\right]-\left[27\right]$ $=0$

Specify the icon for the new element.

The mass number is  $1$ and the atomic number is zero, so the symbol is  ${}_0{}^{1}n$.

The complete nuclear reaction is: 

${}_0{}^{1}n+{}_{27}{}^{59}\mathrm{Co}\to {}_{25}{}^{56}\mathrm{Mn}+{}_2{}^4\mathrm{He}$

Step 4- Given information (part b)

Beta particle ($\mathrm{\beta }$) When emitted from an element, its nucleus converts neutrons into protons and emits electrons.  

Beta $\left({}_{-1}{}^{0}e\right)$ Particles have a negative charge. When  beta particles are released from the nucleus, their mass numbers remain constant.

The incomplete core equation is:  

$?\to {}_7{}^H \mathrm{N}+{}_{-1}{}^{e}e$

Find the missing mass number in the above equation.

$\text{ Mass number of old nucleus = mass number of nitrogen nucleus }$$+$ mass number of beta particle

Substitute the value of the above expression. 

$Mass number of old nucleus =14-0$$=14$

Step 5- Explanation (part b)

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

$atomic number of old nucleus = atomic number of nitrogen nucleus$$+ atomic number of beta particle$

Substitute the value of the above expression 

$atomic number of old nucleus =7-1$$=6$.

Determine the symbol for the new element. 

In the periodic table, atomic number $6$ represents carbon $( C)$.

The core of this  $\left(C\right)$isotope  is written as ${}_6{}^{14}\mathrm{C}\text{. }$ 

Complete the nuclear reaction  as follows. 

${}_6{}^{14}\mathrm{C}\to {}_7{}^{14} \mathrm{N}+{}_{-1}{}^{0}e$

Step 6- Given information (part c)

 

 The incomplete core equation is:

${}_{-1}{}^{6}e+{}_{36}{}^{76}\mathrm{Kr}\to ?$

Find the missing mass number in the above equation.

$\text{ Mass number of new nucleus = mass number of }\mathrm{Kr}+\text{ mass number of beta particle }$

Substitute the value of the above expression.  

$Mass number of new nucleus =76+0$$=76$

Step 7- Explanation (part c)

 Find the missing atomic number in the above equation. 

$\text{ atomic number of new nucleus = atomic number of }\mathrm{Kr}+\text{ atomic number of beta particle }$

Substitute the value of the above expression 

$Atomic number of new nucleus =36-1$$=35$

Determine the symbol for the new element. 

In the periodic table, atomic number $35$ is for bromine $\left(\mathrm{Br}\right)$
.

The core of this $( Br)$ isotope is described as ${}_{35}{}^{76}\mathrm{Br}\text{. }$

 The complete nuclear reaction is:

${}_{-1}{}^{6}e+{}_{36}{}^{76}\mathrm{Kr}\to {}_{35}{}^{76}\mathrm{Br}$

Step 8- Given information (part d)

The incomplete nuclear reactions are: 

${}_2{}^4\mathrm{He}+{}_{85}{}^{241}\mathrm{Am}\to ?+2{}_{\mathrm{e}}{}^1\mathrm{n}$

Find the mass number of the missing element. 

$\text{ Mass number of new nucleus }=\left[\right(\text{ mass number of He })+(\text{ mass number of Am }\left)\right]-$

                                                         $\left[\text{ (mass number of neutrons) }\right]$

Substitute the value of the above expression.  

$Mass number of new nucleus =\left[\right(4)+(241\left)\right]-\left[\right(2\left)\right]$ $=243$

Step 9- Explanation (part d)

$\text{ Atomic number of new nucleus }=\left[\right(\text{ atomic number of He })+(\text{ atomic number of Am }\left)\right]-$                                                                            $\left[\right(\text{ atomic number of neutrons }\left)\right]$

Substitute the value of the above expression 

$Atomic number of new nucleus =\left[\right(2)+(95\left)\right]-\left[\right(0\left)\right]$ $=97$

Determine the symbol for the new element.

In the periodic table, atomic number $97$ represents berkelium (Bk).  

The core of this isotope  is written as

The complete nuclear reaction is:

${}_2{}^4\mathrm{He}+{}_{95}{}^{241}\mathrm{Am}\to {}_{97}{}^{243}\mathrm{Bk}+2{}_0{}^{1}n$