Question

Complete the reaction ${ }_{0} \mathrm{n}^{1}+{ }_{92} \mathrm{U}^{235} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}+3\left({ }_{0} \mathrm{n}^{1}\right)$ (A) \(_{36} \mathrm{Kr}^{90}\) (B) \(_{36} \mathrm{Kr}^{89}\) (C) \(_{36} \mathrm{Kr}^{91}\) (D) \(_{36} \mathrm{Kr}^{92}\)

Short answer

The correct product of the nuclear reaction is (B) \(_{36} \mathrm{Kr}^{89}\).

Step-by-step solution

Identify the initial masses and atomic numbers

: For the given reaction, we have an initial mass and atomic number for each particle involved: \(\mathrm{n}^{1}({ }_{0})+\mathrm{U}^{235}({ }_{92}) \rightarrow \mathrm{Ba}^{144}({ }_{56})+\mathrm{X}^{\mathrm{A}}({ }_{\mathrm{Z}})+3\left(\mathrm{n}^{1}({ }_{0})\right)\) Mass numbers are represented by the superscripts (A), and atomic numbers are represented by the subscripts (Z).

Determine the conservation laws

: According to the conservation of mass number (A), the sum of the initial mass numbers must equal the sum of the final mass numbers: A_initial = A_final For atomic numbers: Z_initial = Z_final

Apply mass number conservation

: Using the mass number conservation, we have: A_initial = 1 (neutron) + 235 (Uranium) = A_final A_final = 144 (Barium) + A (product) + 3 (neutrons) Combining terms, we get: A = (1 + 235) - (144 + 3) = 236 - 147 = 89

Apply atomic number conservation

: Using the atomic number conservation, we have: Z_initial = 0 (neutron) + 92 (Uranium) = Z_final Z_final = 56 (Barium) + Z (product) + 0 (neutrons) Combining terms, we get: Z = (0 + 92) - (56 + 0) = 92 - 56 = 36

Identify the correct product

: Now that we found the mass number A = 89 for the product, and atomic number Z = 36, we can identify the correct product: (B) \(_{36} \mathrm{Kr}^{89}\)