Question:

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

Updated On: Nov 27, 2024
  • Proton
  • Electron
  • Neutron
  • \(\gamma\)-radiations
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The Correct Option is C

Solution and Explanation

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 (\(R = \text{neutrons}\)).

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