The disintegration energy $$Q$$ for the nuclear fission of $$^{235}U \rightarrow ^{140}Ce + ^{94}Zr + n$$ 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 $$c^2 = 931$$ MeV/u

Find the disintegration energy $$Q$$ for $$^{235}U \rightarrow ^{140}Ce + ^{94}Zr + n$$.

$$Q = (\text{mass of reactants} - \text{mass of products}) \times c^2$$

In atomic mass units with $$c^2 = 931$$ MeV/u:

$$Q = (M_U - M_{Ce} - M_{Zr} - M_n) \times 931 \text{ MeV}$$

$$Q = (235.0439 - 139.9054 - 93.9063 - 1.0086) \times 931$$

$$\Delta m = 235.0439 - 139.9054 - 93.9063 - 1.0086$$

$$= 235.0439 - 234.8203 = 0.2236 \text{ u}$$

$$Q = 0.2236 \times 931 = 208.17 \approx 208 \text{ MeV}$$

The correct answer is 208 MeV.