Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

The number of neutrons accompanying the formation of \(_{54} \mathrm{Xe}^{139}\) and \({ }_{38} \mathrm{Sr}^{94}\) from the absorption of slow neutrons by \({ }_{92} \mathrm{U}^{235}\) followed by nuclear fission is (a) 0 (b) 2 (c) 1 (d) 3

Step-by-Step Solution

Answer

3 neutrons are produced.

Step 1: Write the fission equation

Start by writing the nuclear fission equation for the absorption of a neutron by \( {}_{92} \mathrm{U}^{235} \), which leads to the formation of certain fission fragments: \[ _{92}^A\mathrm{U} + _{0}^1\mathrm{n} \rightarrow \, _{54} \mathrm{Xe}^{139} + \, _{38} \mathrm{Sr}^{94} + x \cdot _{0}^{1}\mathrm{n} \] Here, \( x \) represents the number of neutrons produced.

Step 2: Conserve the mass number

Apply conservation of mass number. The mass numbers of the reactant and products should equal: \[ 235 + 1 = 139 + 94 + x \cdot 1 \]Simplify the equation to find \( x \).

Step 3: Solve for the number of produced neutrons

Simplify the equation from Step 2:\[ 236 = 233 + x \] Subtract to find \( x \):\[ x = 3 \].

Step 4: Conserve the atomic number

Check atomic number conservation to verify the solution. The atomic numbers should also balance:\[ 92 = 54 + 38 \]The atomic numbers are consistent, confirming \( x = 3 \) is correct.

Key Concepts

Neutron Production

Nuclear fission often releases additional neutrons when a heavy nucleus splits into smaller fragments. For example, when a slow neutron is absorbed by uranium-235 ( \( _{92} \mathrm{U}^{235} \) ) and the nucleus undergoes fission, it typically produces fragments, like \( _{54} \mathrm{Xe}^{139} \) and \( _{38} \mathrm{Sr}^{94} \).
The reaction also emits free neutrons, which can go on to initiate further fission reactions, sustaining a chain reaction. Understanding the number of neutrons produced in such reactions is crucial for applications in nuclear power generation and nuclear weaponry.
In our case, during the fission of \( _{92} \mathrm{U}^{235} \) , solving the reaction equation shows that three neutrons ( \( 3 \cdot _{0}^{1} \mathrm{n} \) ) are released.

Mass Number Conservation

Mass number conservation is a fundamental principle in nuclear reactions, stipulating that the total mass number (nucleons) before and after a nuclear reaction must be equal. In fission reactions, like that of \( _{92} \mathrm{U}^{235} \), the mass number is the sum of protons and neutrons in the nucleus and must be conserved across the reactants and products.
In the fission of \( _{92} \mathrm{U}^{235} \) , we start with this nucleus absorbing a single neutron, giving us a total mass number of 236 ( 235 + 1 ).
The equation becomes:

• Initial mass number: \( 236 \) (from \( _{92} \mathrm{U}^{235} \) and one neutron)
• Mass number of products: \( 139 + 94 + 3 \) (including the three emitted neutrons)

This conservation is mathematically confirmed by the equation 236 = 139 + 94 + 3, validating our neutron output.

Atomic Number Conservation

Conservation of the atomic number asserts that the sum of protons in the nucleus must remain constant before and after the nuclear fission process. The atomic number determines the element of an atom and shows that no protons are lost or gained in the reaction.
This principle is crucial for validating reactions because it ensures the identities of elements formed are accurate, reaffirming the fixed nature of barrel counts at the atomic level.
Consider the fission of \( _{92} \mathrm{U}^{235} \). The starting atomic number is 92 from uranium ( \( _{92} \mathrm{U}^{235} \) ), while the resulting fragments hold atomic numbers totaling 92:

• Xenon: \( _{54} \mathrm{Xe}^{139} \), contributing 54
• Strontium: \( _{38} \mathrm{Sr}^{94} \), contributing 38

Summing these numbers, \( 54 + 38 = 92 \), corroborates the conservation, confirming the stability and precision of the reaction's atomic number.