Chapter 11, Problem 12P

To determine

The other nuclei formed apart from neutron particle when iron − 58 and bismuth - 209 undergoes a nuclear fusion reaction?

Answer to Problem 12P

The other nuclei formed apart from neutron particle when iron − 58 and bismuth − 209 undergoes a nuclear fusion reaction is $\text{M}\text{t}$ and the reaction is given by $\text{F}\text{e + }\text{B}\text{i}\to \text{M}\text{t +}\text{n}$.

Explanation of Solution

Given info:

Number of protons in $\text{F}\text{e}$

= 26

Mass Number of $\text{F}\text{e}$

= 58

Number of protons in $\text{B}\text{i}$

= 83

Mass Number of $\text{B}\text{i}$

= 209

Number of protons in neutron particle

= 0

Mass Number of neutron particle

= 1.

Formula used:

Atomic number and mass number before and after nuclear reaction remains the same i.e. $\begin{array}{l}{\text{Z}}_{\text{Reactant}}={\text{Z}}_{\text{Product}}\\ {\text{A}}_{\text{Reactant}}={\text{A}}_{\text{Product}}\end{array}$

Where, Z is the atomic number and A is the mass number.

Calculation:

According to the nuclear fusion of iron − 58 and bismuth - 209 the reaction can be represented as,

$\text{F}\text{e + }\text{B}\text{i}\to \text{X}\text{ +}\text{n}$

Total number of protons (equal to the atomic number) of reactants are,

$\begin{array}{l}{\text{Z}}_{\text{Reactant}}={\text{ Z}}_{\text{F}\text{e}}{\text{ + Z}}_{\text{B}\text{i}}\\ {\text{Z}}_{\text{Reactant}}=\text{ 26}+83\\ {\text{Z}}_{\text{Reactant}}=\text{ 109}\end{array}$

Total number of protons (equal to the atomic number) of products are,

$\begin{array}{l}{\text{Z}}_{\text{Product}}={\text{ Z}}_{\text{n}}\text{ + z}\\ {\text{Z}}_{\text{Product}}=\text{ 0 + z}\end{array}$

Substituting the values in formula,

$\begin{array}{l}{\text{Z}}_{\text{Reactant}}={\text{ Z}}_{\text{Product}}\\ \text{109 = 0 + z}\\ \text{109 = z}\end{array}$

Hence the atomic number of the daughter nuclei is 109.

Total mass number of reactants are,

$\begin{array}{l}{\text{A}}_{\text{Reactant}}{\text{= A}}_{\text{F}\text{e}}{\text{ + A}}_{\text{B}\text{i}}\\ {\text{A}}_{\text{Reactant}}\text{= 58+209}\\ {\text{A}}_{\text{Reactant}}\text{= 267}\end{array}$

Total mass number of products are,

$\begin{array}{l}{\text{A}}_{\text{Product}}{\text{= A}}_{\text{n}}\text{ + a }\\ {\text{A}}_{\text{Product}}\text{= 1 + a}\end{array}$

Substituting the values in formula,

$\begin{array}{l}{A}_{\text{Reactant}}={\text{ A}}_{\text{Product}}\\ \text{267 = 1 + a}\\ \text{266 = a}\end{array}$

Hence the mass number of the daughter nuclei is 266.

Therefore, the daughter nuclei is $\text{M}\text{t}$

The reaction thus can be represented as,

$\text{F}\text{e + }\text{B}\text{i}\to \text{M}\text{t +}\text{n}$.

Conclusion:

Thus, the other nuclei formed apart from neutron particle when iron − 58 and bismuth − 209 undergoes a nuclear fusion reaction is $\text{M}\text{t}$ and the reaction is given by $\text{F}\text{e + }\text{B}\text{i}\to \text{M}\text{t +}\text{n}$.