Analyze the Fission Reaction:
The nuclear fission of \(^ {235}_{92} \text{U}\) typically occurs when it absorbs a neutron, forming \(^ {236}_{92} \text{U}\) in an excited state, which then undergoes fission.
The reaction can be written as:
\[ ^{235}_{92} \text{U} + ^{1}_{0} \text{n} \rightarrow \text{fission fragments} + \text{neutrons} \]
Check for Conservation of Mass Number and Atomic Number:
For each option, verify that the total mass number (A) and atomic number (Z) on the right side of the reaction matches the total on the left side.
Total Mass Number (A): \(235 + 1 = 236\)
Total Atomic Number (Z): \(92\)
Evaluate Each Option:
Option 1: \(^{144}_{56} \text{Ba} + ^{89}_{36} \text{Kr} + 4 ^{1}_{0} \text{n}\)
Mass number: \(144 + 89 + 4 \times 1 = 236\)
Atomic number: \(56 + 36 + 4 \times 0 = 92\)
This satisfies the mass and atomic number balance.
Option 2: \(^{140}_{56} \text{Xe} + ^{94}_{38} \text{Sr} + 3 ^{1}_{0} \text{n}\)
Mass number: \(140 + 94 + 3 \times 1 = 237\)
Atomic number: \(56 + 38 + 3 \times 0 = 94\)
This does not satisfy the balance.
Option 3: \(^{153}_{51} \text{Sb} + ^{99}_{41} \text{Nb} + 3 ^{1}_{0} \text{n}\)
Mass number: \(153 + 99 + 3 \times 1 = 256\)
Atomic number: \(51 + 41 + 3 \times 0 = 92\)
This does not satisfy the balance.
Option 4: \(^{144}_{56} \text{Ba} + ^{89}_{36} \text{Kr} + 3 ^{1}_{0} \text{n}\)
Mass number: \(144 + 89 + 3 \times 1 = 236\)
Atomic number: \(56 + 36 + 3 \times 0 = 92\)
This satisfies the mass and atomic number balance.
Conclusion:
Only Option 4 satisfies the conservation of both mass number and atomic number in the nuclear fission process.
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 : \(A_2 \rightleftharpoons 2A\)