In a hypothetical fission reaction
²³⁶₉₂X → ¹⁴¹₅₆Y + ⁹²₃₆Z + 3R
The identity of emitted particles (R) is:

Question

In a hypothetical fission reaction ²³⁶₉₂X → ¹⁴¹₅₆Y + ⁹²₃₆Z + 3R The identity of emitted particles (R) is:

G1 Physics · Accepted Answer

Atomic number on left side: Z(left) = 92 Atomic number on right side: 56 + 36 = 92 So R has zero charge. Mass number on left side: A(left) = 236 Mass number already present on right side: 141 + 92 = 233 Remaining mass number: 236 - 233 = 3 Thus three neutral particles of mass number 1 are emitted. Therefore, R is neutron.