Chapter 20, Problem 54E

Interpretation Introduction

Interpretation:

The equation for the reaction when ${}_{92}^{235}\text{U}$ is bombarded by a single neutron is to be stated.

Concept introduction:

The elements whose nucleus decays spontaneously are known as radioactive elements. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays. All the elements which have an atomic number greater than $83$ are radioactive. The process by which a radioactive nucleus splits to give two nuclei of smaller masses is known as nuclear fission.

Answer to Problem 54E

The equation for the reaction when ${}_{92}^{235}\text{U}$ is bombarded by a single neutron is shown below.

${}_{92}^{235}\text{U}+{}_0^1\text{n}{\to }_{55}^{144}\text{Cs}+{}_{37}^{90}\text{Rb}+2{}_0^1\text{n}$

Explanation of Solution

When the reactant is bombarded by a single neutron, ${}_0^1\text{n}$, the mass number increases.

The total mass of the reactants is calculated as shown below.

$\text{Total}\text{mass}\text{of}\text{the}\text{reactants}=\text{Mass}\text{of}{}_{\text{92}}^{\text{235}}\text{U}+\text{Mass}\text{of}{}_{\text{0}}^{\text{1}}\text{n}$…(1)

The mass number of ${}_{92}^{235}\text{U}$ is $235$.

The mass number of ${}_0^1\text{n}$ is $1$.

Substitute the mass of ${}_{92}^{235}\text{U}$ and ${}_0^1\text{n}$ in equation (1).

$\begin{array}{rll}\text{Total}\text{mass}\text{of}\text{the}\text{reactants}& =& \text{Mass}\text{of}{}_{\text{92}}^{\text{235}}\text{U}+\text{Mass}\text{of}{}_{\text{0}}^{\text{1}}\text{n}\\ & =& \text{235}+\text{1}\\ & =& \text{236}\end{array}$

During this fission, ${}_{55}^{144}\text{Cs}$ and ${}_{37}^{90}\text{Rb}$ are formed.

The total mass of the products is calculated as shown below.

$\text{Total}\text{mass}\text{of}\text{the}\text{products}=\text{Mass}\text{of}{}_{55}^{144}\text{Cs}+\text{Mass}\text{of}{}_{37}^{90}\text{Rb}$…(2)

The mass number of ${}_{55}^{144}\text{Cs}$ is $144$.

The mass number of ${}_{37}^{90}\text{Rb}$ is $90$.

Substitute the mass of ${}_{55}^{144}\text{Cs}$ and ${}_{37}^{90}\text{Rb}$ in equation (1).

$\begin{array}{rll}\text{Total}\text{mass}\text{of}\text{the}\text{products}& =& \text{Mass}\text{of}{}_{55}^{144}\text{Cs}+\text{Mass}\text{of}{}_{37}^{90}\text{Rb}\\ & =& 144+\text{90}\\ & =& \text{234}\end{array}$

The total mass of the emitted neutrons is calculated as shown below.

$\begin{array}{rll}\text{Total}\text{mass}\text{of}\text{emitted}\text{neutrons}& =& \text{Total}\text{mass}\text{of}\text{products}-\text{Total}\text{mass}\text{of}\text{reactants}\\ & =& \text{236-234}\\ & =& \text{2}\end{array}$

The mass number of $1$ neutron, ${}_0^1\text{n}$ is $1$.

The mass number of $2$ neutrons, ${}_0^1\text{n}$ is $2$.

Therefore, two neutrons will be emitted during the fission reaction. The reaction is written as shown below.

${}_{92}^{235}\text{U}+{}_0^1\text{n}{\to }_{55}^{144}\text{Cs}+{}_{37}^{90}\text{Rb}+2{}_0^1\text{n}$

Conclusion

The equation for the reaction when ${}_{92}^{235}\text{U}$ is bombarded by a single neutron has been rightfully stated.