Consider the following nuclear decay: (initially `.^(236)U_(92)` is at rest)
`._(92)^(236)rarr_(90)^(232)ThrarrX`
If the uranium is at rest before its decay, which one of the following statement is true concerning the final nuclei ?

Options

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

Answer

To solve the problem regarding the nuclear decay of Uranium-236 into Thorium-232 and another particle X, we need to analyze the conservation of momentum and kinetic energy during the decay process.

### Step-by-Step Solution:

1. **Identify the Initial State:**
- The initial nucleus is Uranium-236, denoted as \( _{92}^{236}U \).
- It is stated that Uranium is at rest before the decay, meaning its initial momentum is zero.

2. **Write the Decay Equation:**
- The decay can be represented as:
\[ _{92}^{236}U \rightarrow _{90}^{232}Th + X \]
- Here, \( X \) represents the particle emitted during the decay.

3. **Apply Conservation of Momentum:**
- Since the initial momentum is zero (Uranium is at rest), the total momentum after the decay must also be zero:
\[ p_{Th} + p_X = 0 \]
- This implies that the momentum of Thorium \( (p_{Th}) \) is equal in magnitude and opposite in direction to the momentum of particle \( X \) \( (p_X) \):
\[ p_{Th} = -p_X \]

4. **Relate Momentum to Kinetic Energy:**
- The momentum \( p \) of an object is related to its mass \( m \) and velocity \( v \) by the equation:
\[ p = mv \]
- The kinetic energy \( KE \) of an object is given by:
\[ KE = \frac{1}{2} mv^2 \]
- We can express kinetic energy in terms of momentum:
\[ KE = \frac{p^2}{2m} \]

5. **Analyze the Masses:**
- In this decay, Thorium-232 has a greater mass than particle \( X \) (which is likely an alpha particle).
- Since \( X \) has a smaller mass, for equal momentum, it will have a higher velocity.

6. **Compare Kinetic Energies:**
- Since both particles have equal momentum, we can compare their kinetic energies:
- For Thorium:
\[ KE_{Th} = \frac{p^2}{2m_{Th}} \]
- For particle \( X \):
\[ KE_X = \frac{p^2}{2m_X} \]
- Given that \( m_{Th} > m_X \), it follows that:
\[ KE_X > KE_{Th} \]
- Therefore, particle \( X \) has more kinetic energy than Thorium.

7. **Conclusion:**
- The final conclusion is that the momenta of Thorium and particle \( X \) are equal in magnitude (but opposite in direction), while the kinetic energy of particle \( X \) is greater than that of Thorium.

### Final Answer:
The correct statement is that the momenta have equal magnitudes, but particle \( X \) has more kinetic energy than Thorium.