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:
\(^{92}X^{236} \rightarrow ^{56}Y^{141} + ^{36}Z^{92} + 3R\)
Therefore, the correct answer is the particle "Neutron." This identification fits with common fission reactions where neutrons are usually emitted.
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 (\(R = \text{neutrons}\)).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

In a nuclear fission process, a high mass nuclide (A ≈ 236) with binding energy 7.6 MeV/Nucleon dissociated into middle mass nuclides (A ≈ 118), having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be ____ MeV.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)