A nucleus of mass 220 amu in the free state decays to emit an `alpha`-particle . Kinetic energy of the `alpha`-particle emitted is 5.4 MeV. The recoil energy of the daughter nucleus is

Options

A. 0.25 MeV
B. 0.5 MeV
C. 0.1 MeV
D. 0.75 MeV

Answer

To find the recoil energy of the daughter nucleus after the emission of an alpha particle, we can use the principles of conservation of momentum and the relationship between kinetic energy and momentum.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a nucleus with a mass of 220 amu that decays and emits an alpha particle. The kinetic energy (KE) of the emitted alpha particle is given as 5.4 MeV.
- We need to find the recoil energy of the daughter nucleus after the emission.

2. **Initial Momentum**:
- Since the nucleus is initially at rest, its initial momentum is zero. Thus, the total momentum before the decay is zero.

3. **Conservation of Momentum**:
- After the decay, the momentum of the alpha particle and the daughter nucleus must be equal in magnitude and opposite in direction to conserve momentum.
- Let \( P_\alpha \) be the momentum of the alpha particle and \( P_d \) be the momentum of the daughter nucleus. We have:
\[ P_\alpha + P_d = 0 \implies P_d = -P_\alpha \]

4. **Relating Momentum to Kinetic Energy**:
- The momentum \( P \) of a particle is related to its kinetic energy (KE) by the formula:
\[ P = \sqrt{2m \cdot KE} \]
- For the alpha particle:
\[ P_\alpha = \sqrt{2m_\alpha \cdot KE_\alpha} \]
- For the daughter nucleus:
\[ P_d = \sqrt{2m_d \cdot KE_d} \]

5. **Mass of the Alpha Particle and Daughter Nucleus**:
- The mass of the alpha particle \( m_\alpha \) is approximately 4 amu.
- The mass of the daughter nucleus \( m_d \) is:
\[ m_d = 220 \, \text{amu} - 4 \, \text{amu} = 216 \, \text{amu} \]

6. **Setting Up the Equation**:
- From the conservation of momentum, we have:
\[ P_\alpha = P_d \]
- Substituting the expressions for momentum:
\[ \sqrt{2m_\alpha \cdot KE_\alpha} = \sqrt{2m_d \cdot KE_d} \]

7. **Squaring Both Sides**:
- Squaring both sides gives:
\[ 2m_\alpha \cdot KE_\alpha = 2m_d \cdot KE_d \]
- The factor of 2 cancels out:
\[ m_\alpha \cdot KE_\alpha = m_d \cdot KE_d \]

8. **Substituting Values**:
- Substituting \( m_\alpha = 4 \, \text{amu} \), \( KE_\alpha = 5.4 \, \text{MeV} \), and \( m_d = 216 \, \text{amu} \):
\[ 4 \cdot 5.4 = 216 \cdot KE_d \]

9. **Solving for \( KE_d \)**:
- Rearranging gives:
\[ KE_d = \frac{4 \cdot 5.4}{216} \]
- Calculating:
\[ KE_d = \frac{21.6}{216} = 0.1 \, \text{MeV} \]

10. **Final Answer**:
- The recoil energy of the daughter nucleus is \( 0.1 \, \text{MeV} \).

### Summary:
The recoil energy of the daughter nucleus after the emission of the alpha particle is **0.1 MeV**.