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

The Correct Option is C

To determine the identity of the emitted particles (R) in the given nuclear fission reaction, we need to first understand the principle of nuclear reactions, where both mass numbers and atomic numbers are conserved. The given reaction is:

\(^{92}X^{236} \rightarrow ^{56}Y^{141} + ^{36}Z^{92} + 3R\)

1. Mass number balance: The total mass number on both sides of the reaction must be equal. On the left, the mass number of \(^{92}X^{236}\) is 236. On the right, the total mass number is the sum of mass numbers of \(^{56}Y^{141}\), \(^{36}Z^{92}\), and 3R.
2. Calculating:
   • \(141 + 92 + 3 \cdot A_R = 236\)
   • \(233 + 3 \cdot A_R = 236\)
   • \(3 \cdot A_R = 3\)
   • \(A_R = 1\)
3. Atomic number balance: The total atomic number must also be conserved. The atomic number of \(^{92}X^{236}\) is 92, and for the products, it should add up to 92.
4. Calculating:
   • \(56 + 36 + 3 \cdot Z_R = 92\)
   • \(92 + 3 \cdot Z_R = 92\)
   • \(3 \cdot Z_R = 0\)
   • \(Z_R = 0\)
5. Identity of R: The particle with mass number 1 (\(A_R = 1\)) and atomic number 0 (\(Z_R = 0\)) is the neutron.

Therefore, the correct answer is the particle "Neutron." This identification fits with common fission reactions where neutrons are usually emitted.

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