In the nuclear reaction given below, a neutron hits a uranium nucleus. The values of $a$ and $b$ are respectively

$n + {^{235}_{92}U}→{^{a}_{54}Xe} + {^{94}_{b}Sr} + 2 n$

140 and 38

The correct answer is Option (2) → 140 and 38

Reaction:

$ \; ^1_0n + ^{235}_{92}U \;\;\to\;\; ^a_{54}Xe + ^{94}_bSr + 2 \; ^1_0n $

Conservation of mass number (A):

$1 + 235 = a + 94 + 2(1)$

$236 = a + 96$

$a = 140$

Conservation of atomic number (Z):

$0 + 92 = 54 + b + 0$

$92 = 54 + b$

$b = 38$

Answer: $a = 140, \; b = 38$