Eight or - particles and size B - particles are emitted from an atom of 92238U before it achieves stability. What is the nucleon number of the final product...

Answer Details

The initial atom of Uranium-238 emits eight alpha particles (helium-4 nuclei) and B beta particles (electrons or positrons) in order to reach stability. An alpha particle contains two protons and two neutrons, so the emission of one alpha particle reduces the nucleon number by 4. Thus, the total reduction in nucleon number due to the emission of eight alpha particles is 8 x 4 = 32. A beta particle, on the other hand, has negligible mass and does not affect the nucleon number of the atom. Therefore, the nucleon number of the final product in the chain reaction can be calculated by subtracting the total reduction in nucleon number (32) from the original nucleon number of 238. 238 - 32 = 206 Hence, the nucleon number of the final product is 206. Therefore, the correct option is (a) 206.