• # question_answer                 Complete the equation for the following fission process:                                                                              $_{92}{{U}^{235}}+{{\,}_{0}}{{n}^{1}}\,\xrightarrow[{}]{{}}{{\,}_{38}}S{{r}^{90}}+.......$                A)                 $_{54}X{{e}^{143}}+\,3{{\,}_{0}}{{n}^{1}}$ B)                 $_{54}X{{e}^{145}}$                     C)                 $_{57}X{{e}^{142}}$                     D)                 $_{54}X{{e}^{142}}+{{\,}_{0}}{{n}^{1}}$

Key Idea: In a nuclear reaction, atomic mass and charge number remain conserved.                 For a nuclear reaction to be completed, the mass number and charge number on both sides should be same.                 If we complete the equation by choice (a), then the complete reaction is                 Total atomic number on LHS = 92 + 0 = 92                 $_{92}{{U}^{235}}+{{\,}_{0}}{{n}^{1}}\,\to \,{{\,}_{38}}S{{r}^{90}}+{{\,}_{54}}X{{e}^{143}}+\,3{{\,}_{0}}{{n}^{1}}$                 Total atomic number on RHS = 38 + 54 + 0 = 92                 Total mass number on LHS = 235 + 1 = 236                 Total mass number on $RHS=90+143+3\times 1=236$                 Thus, choice (a) is correct.