MHT CET Medical202625 April 2026Evening ShiftPhysicsNuclear PhysicsActual
Consider the following nuclear fission reaction ₀^1 n + ₉₂²³⁵U _Z^A Y + ₄₁⁹⁹Nb + neutrons . If 4 neutrons are released during the fission process, then A and Z are
Options
- AA = 133 and Z = 51
- BA = 51 and Z = 133
- CA = 339 and Z = 133
- DA = 133 and Z = 339
Correct answer
A. A = 133 and Z = 51
Step-by-step solution
The nuclear reaction is given by: ₀^1 n + ₉₂²³⁵U _Z^A Y + ₄₁⁹⁹Nb + 4 ₀^1 n Applying the principle of conservation of mass number: 1 + 235 = A + 99 + 4(1) 236 = A + 103 A = 133 Applying the principle of conservation of atomic number: 0 + 92 = Z + 41 + 4(0) 92 = Z + 41 Z = 51 Thus, A = 133 and Z = 51 . Answer: A = 133 and Z = 51