Question

Which of the following nuclear fragments corresponding to nuclear fission between neutron $\left( ^1_0 n \right) \, \text{and uranium isotope} \, \left( ^{235}_{92} \text{U} \right) \, \text{is correct:}$

• A. $\frac{144}{56} \text{Ba} + \frac{89}{36} \text{Kr} + 4 \frac{1}{0} \text{n}$
• B. $\frac{140}{56} \text{Xe} + \frac{94}{38} \text{Sr} + 3 \frac{1}{0} \text{n}$
• C. $\frac{153}{51} \text{Sb} + \frac{99}{41} \text{Nb} + 3 \frac{1}{0} \text{n}$
• D. $\frac{144}{56} \text{Ba} + \frac{89}{36} \text{Kr} + 3 \frac{1}{0} \text{n}$

Answer 1

Answer: (D)

$\frac{144}{56} \text{Ba} + \frac{89}{36} \text{Kr} + 3 \frac{1}{0} \text{n}$

Answer 2

Analyze the Fission Reaction:
The nuclear fission of $^ {235}_{92} \text{U}$ typically occurs when it absorbs a neutron, forming $^ {236}_{92} \text{U}$ in an excited state, which then undergoes fission.
The reaction can be written as:
$^{235}_{92} \text{U} + ^{1}_{0} \text{n} \rightarrow \text{fission fragments} + \text{neutrons}$

Check for Conservation of Mass Number and Atomic Number:
For each option, verify that the total mass number (A) and atomic number (Z) on the right side of the reaction matches the total on the left side.
Total Mass Number (A): $235 + 1 = 236$
Total Atomic Number (Z): $92$

Evaluate Each Option:
Option 1: $^{144}_{56} \text{Ba} + ^{89}_{36} \text{Kr} + 4 ^{1}_{0} \text{n}$
Mass number: $144 + 89 + 4 \times 1 = 236$
Atomic number: $56 + 36 + 4 \times 0 = 92$
This satisfies the mass and atomic number balance.

Option 2: $^{140}_{56} \text{Xe} + ^{94}_{38} \text{Sr} + 3 ^{1}_{0} \text{n}$
Mass number: $140 + 94 + 3 \times 1 = 237$
Atomic number: $56 + 38 + 3 \times 0 = 94$
This does not satisfy the balance.

Option 3: $^{153}_{51} \text{Sb} + ^{99}_{41} \text{Nb} + 3 ^{1}_{0} \text{n}$
Mass number: $153 + 99 + 3 \times 1 = 256$
Atomic number: $51 + 41 + 3 \times 0 = 92$
This does not satisfy the balance.

Option 4: $^{144}_{56} \text{Ba} + ^{89}_{36} \text{Kr} + 3 ^{1}_{0} \text{n}$
Mass number: $144 + 89 + 3 \times 1 = 236$
Atomic number: $56 + 36 + 3 \times 0 = 92$
This satisfies the mass and atomic number balance.

Conclusion:
Only Option 4 satisfies the conservation of both mass number and atomic number in the nuclear fission process.