Question

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

• A. Proton
• B. Electron
• C. Neutron
• D. $\gamma$-radiations

Answer 1

Answer: (C)

Neutron

Answer 2

To identify the emitted particles, let us verify the conservation of atomic number ($Z$) and mass number ($A$).

- Atomic number ($Z$):

$Z_{\text{LHS}} = 92, \quad Z_{\text{RHS}} = 56 + 36 = 92$

$Z$ is conserved.

- Mass number ($A$):

$A_{\text{LHS}} = 236, \quad A_{\text{RHS}} = 141 + 92 = 233$

The mass number is not conserved. The difference is:

$A_{\text{LHS}} - A_{\text{RHS}} = 236 - 233 = 3$

The missing mass corresponds to three neutrons ($R = \text{neutrons}$).