given reaction,
92U235+0n1→56Ba141+36Kr92+3x+Q(energy)
∑ Atomic number on L.H.S. =92+0=92
∑ Atomic number on R.H.S. =56+36=92
∴ number of protons on L.H.S. are equal to the number of protons on R.H.S. hence x not contains protons in it.
∑ Atomic mass number on L.H.S. =235+1=236
∑ Atomic mass number on R.H.S =141+92+3x=233+3x
L.H.S=R.H.S
233+3x=236
x=1
Mass number=1, no. of proton =0
therefor particle will be Neutron.