The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n i $s ______$ MeV.
Given atomic masses: ²³⁵U = 235.0439 u, ¹⁴⁰Ce = 139.9054 u, ⁹⁴Zr = 93.9063 u, n = 1.0086 u, c² = 931 MeV/u.

Question

The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n i $s ______$ MeV.

Given atomic masses: ²³⁵U = 235.0439 u, ¹⁴⁰Ce = 139.9054 u, ⁹⁴Zr = 93.9063 u, n = 1.0086 u, c² = 931 MeV/u.

G1 Physics · Accepted Answer

Initial mass: m(reactant) = 235.0439 u Final mass: m(product) = 139.9054 + 93.9063 + 1.0086 = 234.8203 u Mass defect: Δm = 235.0439 - 234.8203 = 0.2236 u Energy released: Q = Δm c² = 0.2236 × 931 ≈ 208.17 MeV Therefore: Q ≈ 208 MeV

The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n i s ______ MeV. Given atomic masses: ²³⁵U = 235.0439 u, ¹⁴⁰Ce = 139.9054 u, ⁹⁴Zr = 93.9063 u, n = 1.0086 u, c² = 931 MeV/u.