___________ + 10n    ⟶  14256Ba +9136Kr +  310nA. 23593Np B. 23592U C. 23692U D. 23693Np

The question seems to be about nuclear fission, specifically the fission of a heavy nucleus into two lighter ones, with the emission of three neutrons.

The equation for nuclear fission is usually written as:

Heavy Nucleus + Neutron ⟶ Fission Product 1 + Fission Product 2 + 3 Neutrons

In this case, the fission products are 14256Ba (Barium-142) and 9136Kr (Krypton-91), and the number of neutrons produced is 3.

We can find the heavy nucleus by adding up the atomic numbers and mass numbers on both sides of the equation.

On the right side, the total atomic number is 56 (from Ba) + 36 (from Kr) = 92. The total mass number is 142 (from Ba) + 91 (from Kr) + 3 (from the three neutrons) = 236.

So, the heavy nucleus must have an atomic number of 92 and a mass number of 236.

Therefore, the correct answer is C. 23692U (Uranium-236).