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

Solution

Physics sir · 4 steps

Find the mass that goes missing in the split, then convert it to energy at 931 MeV per u.

1. Mass before. Only the ²³⁵U atom is on the left side, so the initial mass is 235.0439 u.
2. Mass after. Add all three products, including the neutron: 139.9054 + 93.9063 + 1.0086 = 234.8203 u.
3. Mass defect. Δ m = 235.0439 - 234.8203 = 0.2236 u. The products are lighter, so energy is released.

   Q = Δ m c² = 0.2236 × 931 MeV = 208.17 MeV

4. So. The disintegration energy is about 208 MeV, so the answer is 208.

The common slip: 1147. Leaving the neutron out of the products gives Δ m = 1.2322 u and about 1147 MeV, which is far too large for one fission.

The idea to keep. For any Q-value, take the total mass before minus the total mass after, and count every particle on each side. Then multiply by 931 MeV/u.

Answer from the NTA final answer key. Solution by PrepWiser.