In a hypothetical fission reaction
\(^{92}X^{236} \rightarrow ^{56}Y^{141} + ^{36}Z^{92} + 3R\) The identity of emitted particles (R) is:
\(^{92}X^{236} \rightarrow ^{56}Y^{141} + ^{36}Z^{92} + 3R\) The identity of emitted particles (R) is:
- Proton
- Electron
- Neutron
- \(\gamma\)-radiations
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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When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
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- Mass of neutron \( ^{1}_0n \): \( m(^{1}_0n) = 1.00866 \, \text{u} \)
- Conversion factor: \( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)
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