Complete the following nuclear fission reactions :
a) ${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{56}\text{Ba}{+}^{92}\text{Kr}+3_0^1\text{n}+.......$
b)${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }^{148}\text{La}{+}_{35}^{85}\text{Br}+...{.}_0^1\text{n}+energy$

(a) To find atomic number of krypton
By atomic number conservation,
On L.H.S.
Sum of atomic numbers = Z (U) + Z(n) = 92 + 0 = 92
On R.H.S.
Sum of atomic numbers = Z(Ba) + Z(Kr) + 3Z(n) = 56+ Z(Kr) + 3(O)
=56+Z(Kr)
Since L.H.S. = R.H.S.,
92 = 56 + Z(Kr)
Z(Kr) = 36
To find the mass number of Barium
By mass number conservation
On L.H.S.
Sum of mass numbers = A(U)+ A(n)
= 235 + 1 = 236
On R.H.S.
Sum of mass numbers = A(Ba)+ A(Kr) + 3A(n)
= A(Ba) + 92+ 3(1)
= A(Ba) + 95
Since L.H.S=R.H.S.
236 = A(Ba) + 95
A(Ba) = 141
In nuclear fission the sum of actual masses of product nuclei is less than the sum of actual mass of the parent nucleus and the neutron, so energy must be released.
${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{56}^{141}\text{Ba}{+}_{36}^{92}\text{Kr}+3_0^1\text{n}+energy$
(b) To find atomic number of La
Let the number of neutrons released be x.
On L.H.S.
Sum of atomic numbers = Z (U) + Z(n)
= 92 + 0
=92
On R.H.S.
Sum of atomic numbers
= Z(La) + Z(Br) + x Z(n)
= Z(La) + 35 + x (0)
= 35 + Z(La)
Since L.H.S.=R.H.S.
92 = 35 + Z(La)
Z(La ) = 57
To find the number of neutrons released
By mass number conservation
On L.H.S.
Sum of mass numbers = A (U) + A(n)
= 235 +1
=236
On R.H.S.
Sum of mass numbers = A(La) + A(Br) + x A(n) =148 + 85 + x (1)
= x + 233
Since L.H.S = R.H.S.
236 = x + 233
x = 3
${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{57}^{148}\text{La}{+}_{35}^{85}\text{Br}+3_0^1\text{n}+energy$