| Complete the equation for the following fission process: [AIPMT 1998] |
| \[_{92}{{U}^{235}}+{{\,}_{0}}{{n}^{1}}\,\xrightarrow[{}]{{}}{{\,}_{38}}S{{r}^{90}}+.......\] |
A) \[_{54}X{{e}^{143}}+\,3{{\,}_{0}}{{n}^{1}}\]
B) \[_{54}X{{e}^{145}}\]
C) \[_{57}X{{e}^{142}}\]
D) \[_{54}X{{e}^{142}}+{{\,}_{0}}{{n}^{1}}\]
Correct Answer: A
Solution :
| 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 , 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 |
| \[=90+143+3+1=236\] |
| Thus, choice is correct. |
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