A fission reaction is given by `_(92)^(236) U rarr_(54)^(140) Xe + _(38)^(94)St + x + y` , where x and y are two particle Consider `_(92)^(236) U` to be at rest , the kinetic energies of the products are deneted by `k_(xe) K _(st) K _(x) (2MeV ) and `Ky(2MeV) repectively . Let the binding energy per nucleus of `_(92)^(236) U, _(54)^(140) Xe and _(38)^(94)St be 7.5 MeV , 8.4 MeV and 8.5 MeV, ` respectively Considering different conservation laws, the correct option (s) is (are)

Options

A. a. `X= n,y=n,K_(Sr)=129MeV, K_(xe)=86MeV`
B. b. `x=p, y=e^(-), K_(Sr)=129 MeV, K_(xe)=86 MeV`
C. c. `x=p,y=n,K_(Sr)=129 MeV, K_(xe)=86 MeV`
D. d. `x=n, y=n,K_(Sr)= 86 MeV, K_(xe)=129 MeV`

Explanation

To solve the fission reaction given by \( _{92}^{236} U \rightarrow _{54}^{140} Xe + _{38}^{94} Sr + x + y \), we need to apply the conservation laws of mass number and atomic number, as well as the conservation of energy.

### Step 1: Conservation of Mass Number
The total mass number before the reaction must equal the total mass number after the reaction.

\[ 236 = 140 + 94 + A_x + A_y \]

Where \( A_x \) and \( A_y \) are the mass numbers of particles \( x \) and \( y \).

Calculating the right side:

\[ 236 = 234 + A_x + A_y \]

This simplifies to:

\[ A_x + A_y = 2 \]

### Step 2: Conservation of Atomic Number
The total atomic number before the reaction must equal the total atomic number after the reaction.

\[ 92 = 54 + 38 + Z_x + Z_y \]

Where \( Z_x \) and \( Z_y \) are the atomic numbers of particles \( x \) and \( y \).

Calculating the right side:

\[ 92 = 92 + Z_x + Z_y \]

This simplifies to:

\[ Z_x + Z_y = 0 \]

### Step 3: Analyzing the Results
From the conservation of mass number, we found that \( A_x + A_y = 2 \). From the conservation of atomic number, we found that \( Z_x + Z_y = 0 \).

Since \( Z_x + Z_y = 0 \), both \( Z_x \) and \( Z_y \) must be zero, indicating that both \( x \) and \( y \) are neutral particles. The only neutral particle that fits this description with a mass number of 1 is a neutron.

Thus, we can conclude:

\[ x = n \quad \text{and} \quad y = n \]

### Step 4: Kinetic Energy Considerations
Next, we need to consider the kinetic energies of the products. The binding energies per nucleus are given as follows:

- Binding energy of \( _{92}^{236} U = 7.5 \, \text{MeV} \)
- Binding energy of \( _{54}^{140} Xe = 8.4 \, \text{MeV} \)
- Binding energy of \( _{38}^{94} Sr = 8.5 \, \text{MeV} \)

The change in binding energy (\( \Delta E \)) can be calculated as:

\[ \Delta E = \text{(Initial Binding Energy)} - \text{(Final Binding Energy)} \]

Calculating the initial binding energy:

\[ \text{Initial} = 236 \times 7.5 = 1770 \, \text{MeV} \]

Calculating the final binding energy:

\[ \text{Final} = 140 \times 8.4 + 94 \times 8.5 + 2 \times 0 = 1176 + 799 = 1975 \, \text{MeV} \]

Thus, the change in binding energy is:

\[ \Delta E = 1770 - 1975 = -205 \, \text{MeV} \]

### Step 5: Kinetic Energy of Products
The kinetic energy of the products must equal the change in binding energy (converted to kinetic energy). The total kinetic energy of the products is given as:

\[ K_{Xe} + K_{Sr} + K_x + K_y = -\Delta E \]

Given that \( K_{Xe} = 2 \, \text{MeV} \) and \( K_{Sr} = 2 \, \text{MeV} \), we can find:

\[ K_{Xe} + K_{Sr} + K_x + K_y = 2 + 2 + K_x + K_y = 4 + K_x + K_y = 205 \, \text{MeV} \]

This indicates that the kinetic energy of the neutrons must account for the remaining energy.

### Conclusion
From the calculations, we find that:

- \( x \) and \( y \) are both neutrons.
- The correct options based on the analysis are \( x = n \) and \( y = n \).