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: Understanding the Concept:
This problem involves balancing a nuclear reaction. In any nuclear reaction, two quantities must be conserved: 1. The total mass number (A), which is the superscript. 2. The total atomic number (Z), which is the subscript (representing charge).

Step 2: Detailed Explanation:
The given nuclear reaction is: \[ ^1_0\text{n} + ^{235}_{92}\text{U} \rightarrow ^{140}_{54}\text{Xe} + ^b_a\text{Sr} + 2(^1_0\text{n}) \] Conservation of Mass Number (Superscript):
The sum of the mass numbers on the left side must equal the sum on the right side. Left side: \(1 + 235 = 236\) Right side: \(140 + b + 2(1) = 142 + b\) Equating the two sides: \[ 236 = 142 + b \] \[ b = 236 - 142 = 94 \] Conservation of Atomic Number (Subscript):
The sum of the atomic numbers on the left side must equal the sum on the right side. Left side: \(0 + 92 = 92\) Right side: \(54 + a + 2(0) = 54 + a\) Equating the two sides: \[ 92 = 54 + a \] \[ a = 92 - 54 = 38 \]

Step 3: Final Answer:
We have found that the atomic number \(a = 38\) and the mass number \(b = 94\). This corresponds to option (A).