Consider the following nuclear decay: (initially `.^(236)U_(92)` is at rest)
`._(92)^(236)rarr_(90)^(232)ThrarrX`
Regarding this nuclear decay select the correct statement:

Options

A. The nucleus `X` may be at rest.
B. The `._(90)^(232)Th` nucleus may be in excited state.
C. The X may have kinetic energy but `._(90)^(232)Th` will be at rest
D. The `Q` valeus is `Delta mc^(2)` where `Delta mc^(2)` where `Delta m` is mass difference of `(._(92)^(236)U and . _(90)^(232))` and `c` is speed of light

Explanation

To solve the problem regarding the nuclear decay of Uranium-236, we will analyze the decay process step by step and evaluate the statements provided.

### Step 1: Identify the decay process
The decay process given is:
\[ \, _{92}^{236}U \rightarrow \, _{90}^{232}Th + X \]
Here, Uranium-236 decays into Thorium-232 and another particle, which we need to identify.

### Step 2: Determine the identity of particle X
In nuclear decay, the mass number (A) and atomic number (Z) must be conserved.
- The initial mass number is 236 (from Uranium).
- The mass number of Thorium is 232.

To find the mass number of particle X:
\[ A_X = A_U - A_{Th} = 236 - 232 = 4 \]

To find the atomic number of particle X:
- The atomic number of Uranium is 92.
- The atomic number of Thorium is 90.

Thus, the atomic number of particle X is:
\[ Z_X = Z_U - Z_{Th} = 92 - 90 = 2 \]

From this, we can conclude that particle X is an alpha particle, which is represented as:
\[ \, _{2}^{4}He \]

### Step 3: Analyze the statements regarding the decay
1. **Statement A**: "The nucleus X may be at rest."
- After the decay, the conservation of momentum must hold. Since Uranium was initially at rest, the total momentum before decay is zero. Therefore, the momentum of Thorium and the alpha particle must be equal and opposite. This means that it is unlikely for X (the alpha particle) to be at rest. Thus, this statement is **incorrect**.

2. **Statement B**: "The nucleus may be in an excited state."
- After the decay, the resulting Thorium nucleus can be in an excited state due to the energy released during the decay process. This statement is **correct**.

3. **Statement C**: "X may have kinetic energy, but Thorium might not have."
- If the alpha particle (X) has kinetic energy, then by conservation of momentum, Thorium must also have kinetic energy. Thus, this statement is **incorrect**.

4. **Statement D**: "The Q value is given by \( Q = \Delta mc^2 \), where \( \Delta m \) is the mass difference of U and Th."
- The Q value represents the energy released in the decay, which is calculated using the mass difference of the initial and final states. However, it must include the mass of both Thorium and the alpha particle, not just Uranium and Thorium. Thus, this statement is **incorrect**.

### Conclusion
The only correct statement regarding the nuclear decay is **Statement B**: "The nucleus may be in an excited state."