Consider the following nuclear decay: `""_(92)^(236)Uto""_(90)^(234)Th+X`
If the uranium is at rest before its decay, which one of the following statements is true concerning the final nuclei?

Options

A. They have equal kinetic energies, but he thorium nucleus has much more momentum.
B. They have momenta of equal magnitudes, but the thorium nucleus has much more kinetic energy.
C. They have equal kinetic energies and momenta of equal magnitudes
D. They have momenta of equal magnitudes, but X has much more kinetic energy

Answer

To solve the problem regarding the nuclear decay of Uranium-236 into Thorium-234 and another particle (X), we need to analyze the conservation laws that apply during this decay process.

### Step-by-Step Solution:

1. **Identify the Decay Process**:
The decay can be represented as:
\[ \text{Uranium-236} \rightarrow \text{Thorium-234} + X \]
Here, \(X\) represents the other particle produced during the decay.

2. **Conservation of Mass Number**:
The mass number (A) must be conserved in nuclear reactions. The mass number of Uranium-236 is 236, and Thorium-234 has a mass number of 234. Thus, we can find the mass number of particle \(X\):
\[ A_{\text{U}} = A_{\text{Th}} + A_X \implies 236 = 234 + A_X \implies A_X = 2 \]
Therefore, \(X\) is a particle with a mass number of 2, which indicates that \(X\) is likely an alpha particle (\(^4_2He\)).

3. **Conservation of Atomic Number**:
The atomic number (Z) must also be conserved. The atomic number of Uranium-236 is 92, and Thorium-234 has an atomic number of 90. Thus, we can find the atomic number of particle \(X\):
\[ Z_{\text{U}} = Z_{\text{Th}} + Z_X \implies 92 = 90 + Z_X \implies Z_X = 2 \]
This confirms that \(X\) is indeed an alpha particle, as it has an atomic number of 2.

4. **Kinetic Energy and Momentum Considerations**:
Since the uranium nucleus is initially at rest, the total momentum before decay is zero. According to the conservation of momentum:
\[ \text{Momentum of Thorium} + \text{Momentum of X} = 0 \]
This implies that the momentum of Thorium and the momentum of \(X\) are equal in magnitude but opposite in direction:
\[ p_{\text{Th}} = -p_X \]

5. **Kinetic Energy Comparison**:
The kinetic energy of a particle is given by the formula:
\[ KE = \frac{p^2}{2m} \]
Since \(p_{\text{Th}} = -p_X\), we can express the kinetic energies in terms of their masses:
\[ KE_{\text{Th}} = \frac{p_{\text{Th}}^2}{2m_{\text{Th}}}, \quad KE_X = \frac{p_X^2}{2m_X} \]
Given that \(m_X\) (mass of alpha particle) is less than \(m_{\text{Th}}\) (mass of Thorium), the kinetic energy of \(X\) will be greater than that of Thorium.

6. **Conclusion**:
Therefore, the correct statement concerning the final nuclei is that they have momenta of equal magnitudes, but the kinetic energy of \(X\) is greater than that of Thorium.

### Final Answer:
The correct statement is: "They have momenta of equal magnitudes, but the kinetic energy of \(X\) is more as compared to Thorium."