A nucleus of mass `218` amu in Free State decays to emit an `alpha`-particle. Kinetic energy of the `beta-`particle emitted is `6.7 MeV`. The recoil energy (in MeV) of the daughter nucleus is

Options

A. `1.0`
B. `0.5`
C. `0.25`
D. `0.125`

Explanation

To solve the problem of finding the recoil energy of the daughter nucleus after the emission of an alpha particle, we can follow these steps:

### Step 1: Understand the decay process
When a nucleus decays to emit an alpha particle, it transforms into a daughter nucleus. The total energy before and after the decay must be conserved.

### Step 2: Identify the masses involved
- The original nucleus has a mass of 218 amu.
- The alpha particle has a mass of approximately 4 amu.
- The mass of the daughter nucleus can be calculated as:
\[ \text{Mass of daughter nucleus} = \text{Mass of original nucleus} - \text{Mass of alpha particle} = 218 \, \text{amu} - 4 \, \text{amu} = 214 \, \text{amu} \]

### Step 3: Apply conservation of momentum
In the decay process, the momentum before and after the decay must be equal. Since the original nucleus is at rest, the momentum of the daughter nucleus and the alpha particle must be equal in magnitude and opposite in direction.

Let:
- \( p_\alpha \) = momentum of the alpha particle
- \( p_d \) = momentum of the daughter nucleus

From conservation of momentum:
\[ p_d = -p_\alpha \]

### Step 4: Relate kinetic energy to momentum
The kinetic energy (K.E.) of a particle is related to its momentum (p) by the equation:
\[ K.E. = \frac{p^2}{2m} \]
where \( m \) is the mass of the particle.

### Step 5: Calculate the momentum of the alpha particle
Given that the kinetic energy of the emitted alpha particle is \( K.E. = 6.7 \, \text{MeV} \), we can express the momentum of the alpha particle as:
\[ p_\alpha = \sqrt{2 m_\alpha K.E.} \]
where \( m_\alpha \) is the mass of the alpha particle.

### Step 6: Calculate the recoil energy of the daughter nucleus
Using the relationship between kinetic energy and momentum for the daughter nucleus:
\[ K.E._d = \frac{p_d^2}{2m_d} \]
Since \( p_d = p_\alpha \), we can substitute:
\[ K.E._d = \frac{p_\alpha^2}{2m_d} \]

### Step 7: Substitute the values
Using the mass of the alpha particle \( m_\alpha = 4 \, \text{amu} \) and the mass of the daughter nucleus \( m_d = 214 \, \text{amu} \):
\[ K.E._d = \frac{(p_\alpha)^2}{2 \times 214} \]

Now, substituting for \( p_\alpha \):
\[ K.E._d = \frac{(2 \times 4 \times 6.7 \, \text{MeV})^2}{2 \times 214} \]

### Step 8: Calculate the recoil energy
After performing the calculations:
\[ K.E._d \approx 0.125 \, \text{MeV} \]

### Final Answer
The recoil energy of the daughter nucleus is approximately \( 0.125 \, \text{MeV} \).

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