In a hypothetical fission reaction $$_{92}X^{236} \rightarrow _{56}Y^{141} + _{36}Z^{92} + 3R$$. The identity of emitted particles (R) is :

We are given the hypothetical fission reaction:

$$_{92}X^{236} \rightarrow \, _{56}Y^{141} + \, _{36}Z^{92} + 3R$$

We need to identify the emitted particle $$R$$.

Applying conservation of mass number (A), in any nuclear reaction the total mass number must be conserved (total on left = total on right). On the left side $$A = 236$$, and on the right side $$141 + 92 + 3A_R = 233 + 3A_R$$. Setting them equal gives $$236 = 233 + 3A_R$$, hence $$3A_R = 3$$ and $$A_R = 1$$, so each particle $$R$$ has mass number 1.

Applying conservation of atomic number (Z), the total charge (atomic number) must also be conserved. On the left side $$Z = 92$$, and on the right side $$56 + 36 + 3Z_R = 92 + 3Z_R$$. Equating these yields $$92 = 92 + 3Z_R$$, so $$3Z_R = 0$$ and $$Z_R = 0$$. Each particle $$R$$ therefore has atomic number 0 (no charge).

A particle with mass number $$A = 1$$ and atomic number $$Z = 0$$ is a neutron ($$^1_0 n$$). This makes physical sense, as neutron emission is very common in nuclear fission reactions. The emitted neutrons can go on to trigger further fission events, which is the basis of a nuclear chain reaction.

The correct answer is Option (2): Neutron.