what kinetic energy must an `alpha`-particle possess to split a deuteron `H^(2)` whose binding energy is `E_(b)=2.2 MeV`?

Explanation

To find the kinetic energy that an alpha particle must possess to split a deuteron (H²) with a binding energy of \( E_b = 2.2 \, \text{MeV} \), we can follow these steps:

### Step 1: Understand the problem
The deuteron is a bound state of a proton and a neutron, and it has a binding energy of \( 2.2 \, \text{MeV} \). To split the deuteron, the alpha particle must provide enough energy to overcome this binding energy.

### Step 2: Set up the conservation of momentum
Let:
- Mass of the alpha particle, \( m_{\alpha} = 4m \) (where \( m \) is the mass of a nucleon)
- Mass of the deuteron, \( m_d = 2m \)
- Initial velocity of the deuteron, \( u_d = 0 \)
- Initial velocity of the alpha particle, \( v_{\alpha} = v \)

According to the conservation of linear momentum:
\[ m_{\alpha} v_{\alpha} + m_d u_d = (m_{\alpha} + m_d) v' \]
Substituting the values:
\[ 4m \cdot v + 2m \cdot 0 = (4m + 2m) v' \]
This simplifies to:
\[ 4mv = 6mv' \]

### Step 3: Solve for the final velocity
From the equation above, we can solve for \( v' \):
\[ v' = \frac{4}{6} v = \frac{2}{3} v \]

### Step 4: Calculate the change in kinetic energy
The initial kinetic energy of the alpha particle is:
\[ KE_{\text{initial}} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 = \frac{1}{2} (4m) v^2 = 2mv^2 \]
After the interaction, the total mass is \( 6m \) and the final velocity is \( v' = \frac{2}{3} v \):
\[ KE_{\text{final}} = \frac{1}{2} (6m) \left(\frac{2}{3} v\right)^2 = \frac{1}{2} (6m) \left(\frac{4}{9} v^2\right) = \frac{24}{18} mv^2 = \frac{4}{3} mv^2 \]

### Step 5: Calculate the change in kinetic energy
The change in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 2mv^2 - \frac{4}{3} mv^2 \]
Finding a common denominator:
\[ \Delta KE = \frac{6}{3} mv^2 - \frac{4}{3} mv^2 = \frac{2}{3} mv^2 \]

### Step 6: Relate the change in kinetic energy to the binding energy
For the alpha particle to split the deuteron, the change in kinetic energy must be equal to or greater than the binding energy:
\[ \frac{2}{3} mv^2 \geq E_b \]
Substituting \( E_b = 2.2 \, \text{MeV} \):
\[ \frac{2}{3} mv^2 \geq 2.2 \, \text{MeV} \]
Thus, the minimum kinetic energy \( KE_{\text{min}} \) required is:
\[ mv^2 \geq 3.3 \, \text{MeV} \]

### Step 7: Calculate the required kinetic energy of the alpha particle
Since the mass of the alpha particle is \( 4m \):
\[ KE_{\alpha} = \frac{1}{2} (4m) v^2 = 2mv^2 \]
From the previous inequality:
\[ 2mv^2 \geq 6.6 \, \text{MeV} \]
Thus, the minimum kinetic energy that the alpha particle must possess is:
\[ KE_{\alpha} \geq 6.6 \, \text{MeV} \]

### Final Answer
The kinetic energy that the alpha particle must possess to split the deuteron is **6.6 MeV**.

---