Question

In the following nuclear reaction, \(^1_0\text{n} + ^{235}_{92}\text{U} \rightarrow ^{140}_{54}\text{Xe} + ^b_a\text{Sr} + 2(^1_0\text{n})\) we have

• A. a = 38, b = 94
• B. a = 94, b = 38
• C. a = 94, b = 40
• D. a = 96, b = 38

Hint:

When balancing nuclear equations, always balance the superscripts (mass numbers) and subscripts (atomic numbers) separately. It's a simple bookkeeping process. Be careful with coefficients, like the "2" in front of the neutron on the product side.

Answer 1

The Correct Option is A

Step 1: Conceptual Foundation: This problem requires the balancing of a nuclear reaction. Key principles dictate that in any nuclear reaction, two fundamental quantities are conserved: 1. The total mass number (A), denoted by the superscript. 2. The total atomic number (Z), indicated by the subscript and representing charge.

Step 2: Analytical Breakdown: The nuclear reaction under consideration is: \[ ^1_0\text{n} + ^{235}_{92}\text{U} \rightarrow ^{140}_{54}\text{Xe} + ^b_a\text{Sr} + 2(^1_0\text{n}) \] Mass Number Conservation (Superscript): The sum of mass numbers on the reactant side must equal the sum on the product side. Left side: \(1 + 235 = 236\). Right side: \(140 + b + 2(1) = 142 + b\). By equating both sides: \[ 236 = 142 + b \] Solving for b yields: \[ b = 236 - 142 = 94 \] Atomic Number Conservation (Subscript): Similarly, the sum of atomic numbers on the reactant side must equal the sum on the product side. Left side: \(0 + 92 = 92\). Right side: \(54 + a + 2(0) = 54 + a\). By equating both sides: \[ 92 = 54 + a \] Solving for a yields: \[ a = 92 - 54 = 38 \]

Step 3: Conclusion: The calculated values are \(a = 38\) for the atomic number and \(b = 94\) for the mass number. This corresponds to option (A).