A proton collides with a stationary deuteron to form a `.^3He` nucleus. For this reaction to take place , the proton must have a minimum kinetic energy `K_0` . If instead , a deuteron collides with a stationary proton to make a `.^3He` nucleus , then it must have minimum kinetic energy equal to

Options

A. `2K_0`
B. `1.5K_0`
C. `K_0`
D. `K_0/2`

Answer

To solve the problem, we need to analyze the kinetic energy required for two different collision scenarios involving a proton and a deuteron.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- In the first scenario, a proton (p) collides with a stationary deuteron (d) to form a helium-3 nucleus (³He).
- The reaction can be represented as:
\[ p + d \rightarrow \, ^3He \]

2. **Minimum Kinetic Energy for Proton**:
- For the proton to initiate the reaction, it must have a minimum kinetic energy \( K_0 \).
- The kinetic energy of the proton can be expressed as:
\[ K_0 = \frac{1}{2} m_p v^2 \]
- Here, \( m_p \) is the mass of the proton and \( v \) is its velocity.

3. **Analyzing the Second Scenario**:
- In the second scenario, we have a deuteron colliding with a stationary proton.
- The reaction can be represented as:
\[ d + p \rightarrow \, ^3He \]

4. **Mass Relationship**:
- The mass of the deuteron (\( m_d \)) is approximately twice the mass of the proton:
\[ m_d = 2 m_p \]

5. **Kinetic Energy of Deuteron**:
- Let the minimum kinetic energy required for the deuteron in this scenario be \( K' \).
- The kinetic energy of the deuteron can be expressed as:
\[ K' = \frac{1}{2} m_d v'^2 \]
- Substituting \( m_d = 2 m_p \):
\[ K' = \frac{1}{2} (2 m_p) v'^2 \]
\[ K' = m_p v'^2 \]

6. **Relating the Velocities**:
- To find the relationship between the velocities \( v \) (of the proton in the first case) and \( v' \) (of the deuteron in the second case), we note that for the same reaction to occur, the deuteron must have sufficient energy to compensate for its larger mass.
- Since the kinetic energy is proportional to the mass and the square of the velocity, we can relate the two energies:
\[ K' = 2 K_0 \]
- This means that the minimum kinetic energy \( K' \) required for the deuteron is twice that of the proton's minimum kinetic energy \( K_0 \).

### Final Result:
Thus, the minimum kinetic energy \( K' \) required for the deuteron in the second scenario is:
\[ K' = 2 K_0 \]