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

<p>In order to identify the correct nuclear fragments resulting from the fission of uranium-235 by a neutron, $\left({ }_0^1 \mathrm{n}\right) + \left({ }_{92}^{235} \mathrm{U}\right)$, we need to apply the conservation of mass number and atomic number. These conservation laws tell us that the sum of mass numbers (top numbers, representing the total count of protons and neutrons) and the sum of atomic numbers (bottom numbers, representing the total count of protons) before and after the fission must be equal. Let's apply these rules to each option:</p> <p>For the uranium-235 fission reaction, before the reaction, the total mass number is $235 + 1 = 236$ and the total atomic number is $92$ (since a neutron doesn't contribute to the atomic number).</p> <p><strong>Option A:</strong> ${ }_{56}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+3{ }_0^1 \mathrm{n}$ <ul> <li>Total mass number after = $140 + 94 + 3(1) = 234 + 3 = 237$</li> <li>Total atomic number after = $56 + 38 = 94$</li> </ul> </p> <p><strong>Option B:</strong> ${ }_{51}^{153} \mathrm{Sb}+{ }_{41}^{99} \mathrm{Nb}+3{ }_0^1 \mathrm{n}$ <ul> <li>Total mass number after = $153 + 99 + 3(1) = 252 + 3 = 255$</li> <li>Total atomic number after = $51 + 41 = 92$</li> </ul> </p> <p><strong>Option C:</strong> ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+4{ }_0^1 \mathrm{n}$ <ul> <li>Total mass number after = $144 + 89 + 4(1) = 233 + 4 = 237$</li> <li>Total atomic number after = $56 + 36 = 92$</li> </ul> </p> <p><strong>Option D:</strong> ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n}$ <ul> <li>Total mass number after = $144 + 89 + 3(1) = 233 + 3 = 236$</li> <li>Total atomic number after = $56 + 36 = 92$</li> </ul> </p> <p>By examining the conservation of mass numbers and atomic numbers, <strong>Option D</strong> is identified as the correct option because both the total mass number and atomic number after the reaction match exactly with the total mass number and atomic number before the reaction. Therefore, the nuclear fission fragments and the neutron count for the reaction are accurately represented by ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n}$. </p>