A nucleus with mass number 220 initially at rest emits an `alpha`-particle. If the Q-value of the reaction is 5.5 MeV, calculate the kinetic energy of the `alpha`-particle.
(a) 4.4 MeV (b) 5.4 MeV (c) 5.6 MeV (d) 6.5 MeV

Explanation

To solve the problem, we need to find the kinetic energy of the alpha particle emitted from a nucleus with mass number 220, given that the Q-value of the reaction is 5.5 MeV.

### Step-by-Step Solution:

1. **Identify the Mass Numbers**:
- The initial nucleus has a mass number \( A = 220 \).
- When it emits an alpha particle (which has a mass number of 4), the remaining nucleus (let's call it Y) will have a mass number of \( 220 - 4 = 216 \).

2. **Write the Q-value Equation**:
- The Q-value of the reaction is given by the equation:
\[ Q = KE_{\alpha} + KE_Y \]
where \( KE_{\alpha} \) is the kinetic energy of the alpha particle and \( KE_Y \) is the kinetic energy of the remaining nucleus Y.

3. **Use Conservation of Momentum**:
- Since the initial nucleus is at rest, the momentum before emission is zero. Therefore, the momentum after emission must also be zero:
\[ m_{\alpha} v_{\alpha} + m_Y v_Y = 0 \]
where \( m_{\alpha} \) is the mass of the alpha particle, \( v_{\alpha} \) is its velocity, \( m_Y \) is the mass of nucleus Y, and \( v_Y \) is its velocity.

4. **Express \( v_Y \) in terms of \( v_{\alpha} \)**:
- From the momentum conservation, we can express \( v_Y \):
\[ v_Y = -\frac{m_{\alpha}}{m_Y} v_{\alpha} \]

5. **Substitute into the Kinetic Energy Equation**:
- The kinetic energy of the alpha particle is given by:
\[ KE_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]
- The kinetic energy of nucleus Y can be expressed as:
\[ KE_Y = \frac{1}{2} m_Y v_Y^2 = \frac{1}{2} m_Y \left(-\frac{m_{\alpha}}{m_Y} v_{\alpha}\right)^2 = \frac{1}{2} \frac{m_{\alpha}^2}{m_Y} v_{\alpha}^2 \]

6. **Combine the Kinetic Energies**:
- Now substituting back into the Q-value equation:
\[ Q = KE_{\alpha} + KE_Y = \frac{1}{2} m_{\alpha} v_{\alpha}^2 + \frac{1}{2} \frac{m_{\alpha}^2}{m_Y} v_{\alpha}^2 \]
- Factor out \( \frac{1}{2} v_{\alpha}^2 \):
\[ Q = \frac{1}{2} v_{\alpha}^2 \left( m_{\alpha} + \frac{m_{\alpha}^2}{m_Y} \right) \]

7. **Substitute Mass Values**:
- The mass of the alpha particle \( m_{\alpha} \) corresponds to 4 (mass number) and the mass of Y \( m_Y \) corresponds to 216.
- Thus, we have:
\[ Q = \frac{1}{2} v_{\alpha}^2 \left( 4 + \frac{4^2}{216} \right) \]

8. **Solve for \( KE_{\alpha} \)**:
- Rearranging gives:
\[ KE_{\alpha} = Q \cdot \frac{m_Y}{m_Y + m_{\alpha}} = 5.5 \cdot \frac{216}{220} \]
- Calculate:
\[ KE_{\alpha} = 5.5 \cdot \frac{216}{220} = 5.4 \text{ MeV} \]

### Final Answer:
The kinetic energy of the alpha particle is **5.4 MeV**.