In the following nuclear reaction, ${^1_0n} + {^{235}_{92}U} → {^{140}_{54}Xe}+ {^a_bSr}+ 2{^1_0n}$ we have

$a= 94, b = 38$

The correct answer is Option (2) → $a= 94, b = 38$ **

$\text{Given nuclear reaction: } \, ^1_0n + ^{235}_{92}U \rightarrow ^{140}_{54}Xe + ^a_bSr + 2\,^1_0n$

$\text{Balancing mass numbers: } 1 + 235 = 140 + a + 2(1)$

$236 = 140 + a + 2$

$a = 94$

$\text{Balancing atomic numbers: } 0 + 92 = 54 + b + 0$

$92 = 54 + b$

$b = 38$

therefore $\, ^a_bSr = ^{94}_{38}Sr$

$\text{Answer: } a = 94, \, b = 38$