Stationary nucleus `._(92)^(238)U` decays by a emission generating a total kinetic energy T:
`._(92)^(238)U rarr ._(90)^(234)Th +._2^4 alpha`
What is the kinetic energy of the `alpha`-particle?

Options

A. Slightly less than `T//2`
B. `T//2`
C. Slightly less than T
D. Slightly greater than T

Explanation

To solve the problem of finding the kinetic energy of the alpha particle emitted during the decay of a stationary nucleus of uranium-238, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The decay reaction is given as:
\[ \,_{92}^{238}\text{U} \rightarrow \,_{90}^{234}\text{Th} + \,_{2}^{4}\alpha \]
Here, uranium-238 decays into thorium-234 and an alpha particle.

2. **Define Total Kinetic Energy**:
We are given that the total kinetic energy (T) generated during this decay is the sum of the kinetic energies of the thorium nucleus (T_th) and the alpha particle (T_alpha):
\[ T = T_{\text{Th}} + T_{\alpha} \]

3. **Apply Conservation of Momentum**:
Since the uranium nucleus is initially at rest, the total momentum before decay is zero. After the decay, the momentum must also be zero:
\[ 0 = p_{\text{Th}} + p_{\alpha} \]
This implies:
\[ m_{\text{Th}} v_{\text{Th}} + m_{\alpha} v_{\alpha} = 0 \]
Rearranging gives:
\[ m_{\text{Th}} v_{\text{Th}} = - m_{\alpha} v_{\alpha} \]
where \( m_{\text{Th}} \) and \( m_{\alpha} \) are the masses of thorium and the alpha particle, respectively.

4. **Relate Kinetic Energies**:
The kinetic energy of each particle can be expressed as:
\[ T_{\text{Th}} = \frac{1}{2} m_{\text{Th}} v_{\text{Th}}^2 \]
\[ T_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]
Using the momentum relationship, we can express \( v_{\text{Th}} \) in terms of \( v_{\alpha} \):
\[ v_{\text{Th}} = -\frac{m_{\alpha}}{m_{\text{Th}}} v_{\alpha} \]

5. **Substitute and Solve**:
Substitute \( v_{\text{Th}} \) into the kinetic energy expression for thorium:
\[ T_{\text{Th}} = \frac{1}{2} m_{\text{Th}} \left(-\frac{m_{\alpha}}{m_{\text{Th}}} v_{\alpha}\right)^2 = \frac{1}{2} \frac{m_{\alpha}^2}{m_{\text{Th}}} v_{\alpha}^2 \]
Now, substituting this into the total kinetic energy equation:
\[ T = \frac{1}{2} m_{\alpha} v_{\alpha}^2 + \frac{1}{2} \frac{m_{\alpha}^2}{m_{\text{Th}}} v_{\alpha}^2 \]
Factoring out \( \frac{1}{2} v_{\alpha}^2 \):
\[ T = \frac{1}{2} v_{\alpha}^2 \left(m_{\alpha} + \frac{m_{\alpha}^2}{m_{\text{Th}}}\right) \]

6. **Mass Relationship**:
Knowing the masses:
- Mass of alpha particle, \( m_{\alpha} = 4 \) (approximately)
- Mass of thorium, \( m_{\text{Th}} = 234 \)

We can express \( T_{\alpha} \) in terms of \( T \):
\[ T_{\alpha} = \frac{m_{\alpha}}{m_{\alpha} + m_{\text{Th}}} T \]

7. **Final Calculation**:
Substitute the values:
\[ T_{\alpha} = \frac{4}{4 + 234} T = \frac{4}{238} T = \frac{2}{119} T \]
This gives us the kinetic energy of the alpha particle.

### Final Result:
The kinetic energy of the alpha particle is:
\[ T_{\alpha} = \frac{2}{119} T \]