The compound unstabel nucleus `._(92)^(236)U` often decays in accordance with the following reaction
`_(92)^(236)U rarr ._(54)^(140)Xe +_(38)^(94)Sr+` other particles
During the reaction, the uranium nucleus ''fissions'' (splits) into the two smaller nuceli have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons).
Inside a nucleus, the nucleons (protonsa and neutrons)attract each other with a ''strong nuclear'' force. All neutrons exert approxiamtely the same strong nuclear force on each other. This force holds the nuclear are very close together at intranuclear distances.
In the nuclear reaction presented above, the ''other particles'' might be .

Options

A. An alpha particle, which consists of two protons and two neutrons
B. two protons
C. one proton and one neutron
D. two neutrons

Explanation

To solve the problem, we need to analyze the nuclear reaction given and determine what the "other particles" are that are produced during the fission of uranium-236. We will use the conservation of mass number and atomic number to find the solution.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is:
\[ _{92}^{236}U \rightarrow _{54}^{140}Xe + _{38}^{94}Sr + \text{other particles} \]

2. **Conservation of Mass Number**:
The mass number (A) must be conserved in the reaction.
- The mass number of uranium (U) is 236.
- The mass number of xenon (Xe) is 140.
- The mass number of strontium (Sr) is 94.
- Let the mass number of the other particles be \( A \).

Setting up the equation:
\[ 236 = 140 + 94 + A \]
Simplifying this gives:
\[ 236 = 234 + A \implies A = 2 \]

3. **Conservation of Atomic Number**:
The atomic number (Z) must also be conserved.
- The atomic number of uranium (U) is 92.
- The atomic number of xenon (Xe) is 54.
- The atomic number of strontium (Sr) is 38.
- Let the atomic number of the other particles be \( Z \).

Setting up the equation:
\[ 92 = 54 + 38 + Z \]
Simplifying this gives:
\[ 92 = 92 + Z \implies Z = 0 \]

4. **Identifying the Other Particles**:
Now we know that the other particles must have a mass number of 2 and an atomic number of 0. The only particles that fit this description are two neutrons (since neutrons have a mass number of 1 and an atomic number of 0).

5. **Conclusion**:
Therefore, the "other particles" produced in this reaction are:
\[ \text{2 neutrons} \]

### Final Answer:
The other particles might be **two neutrons**.

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