Initially the nucleus of radium 226 is at rest. It decays due to which and `alpha` particle and the nucleus of radon are created. The released energy during the decay is `4.87` Mev, which appears as the kinetic energy of the two resulted particles `[m_(alpha)=4.002"amu",m_("Rn")=222.017"amu"]`
Kinetic energies of `alpha` particle & radon nucleus are respectively

Options

A. `0.09 "Mev", 4.08 "Mev"`
B. `4.78 "Mev", 0.09 "Mev"`
C. `4.08 "Mev", 0.09 "Mev"`
D. `3.68 "Mev", 0.08 "Mev"`

Explanation

To solve the problem of finding the kinetic energies of the alpha particle and the radon nucleus after the decay of radium-226, we can follow these steps:

### Step 1: Understand the conservation of momentum
Initially, the nucleus of radium-226 is at rest, which means the total initial momentum is zero. After the decay, the momentum of the alpha particle and the radon nucleus must also sum to zero. Therefore, we can express this as:

\[ P_{\alpha} + P_{Rn} = 0 \]

This implies that:

\[ P_{\alpha} = -P_{Rn} \]

### Step 2: Relate momentum to kinetic energy
The momentum \( P \) of a particle is related to its mass \( m \) and velocity \( v \) by the equation:

\[ P = mv \]

The kinetic energy \( K \) of a particle can be expressed in terms of momentum as:

\[ K = \frac{P^2}{2m} \]

### Step 3: Set up the equations for kinetic energies
Let \( K_{\alpha} \) be the kinetic energy of the alpha particle and \( K_{Rn} \) be the kinetic energy of the radon nucleus. From the conservation of momentum, we can express the ratio of their momenta:

\[ \frac{P_{\alpha}}{P_{Rn}} = -\frac{m_{Rn}}{m_{\alpha}} \]

### Step 4: Use the relation between kinetic energy and mass
Using the relation for kinetic energy in terms of momentum, we can write:

\[ \frac{K_{\alpha}}{K_{Rn}} = \frac{P_{\alpha}^2/m_{\alpha}}{P_{Rn}^2/m_{Rn}} = \frac{m_{Rn}}{m_{\alpha}} \]

### Step 5: Calculate the mass ratio
Given the masses:
- \( m_{\alpha} = 4.002 \, \text{amu} \)
- \( m_{Rn} = 222.017 \, \text{amu} \)

The mass ratio is:

\[ \frac{m_{Rn}}{m_{\alpha}} = \frac{222.017}{4.002} \approx 55.5 \]

### Step 6: Use the total energy released
The total energy released during the decay is given as \( 4.87 \, \text{MeV} \). This total energy is shared between the kinetic energies of the alpha particle and the radon nucleus:

\[ K_{\alpha} + K_{Rn} = 4.87 \, \text{MeV} \]

### Step 7: Set up the equations for kinetic energies
Let \( K_{\alpha} = x \) and \( K_{Rn} = 4.87 - x \). From the ratio of kinetic energies, we have:

\[ \frac{x}{4.87 - x} = \frac{m_{Rn}}{m_{\alpha}} = 55.5 \]

### Step 8: Solve for \( x \)
Cross-multiplying gives:

\[ x = 55.5 \cdot (4.87 - x) \]

Expanding this:

\[ x = 270.885 - 55.5x \]

Combining like terms:

\[ x + 55.5x = 270.885 \]

\[ 56.5x = 270.885 \]

\[ x \approx \frac{270.885}{56.5} \approx 4.79 \, \text{MeV} \]

### Step 9: Calculate \( K_{Rn} \)
Now, substituting back to find \( K_{Rn} \):

\[ K_{Rn} = 4.87 - K_{\alpha} \approx 4.87 - 4.79 \approx 0.08 \, \text{MeV} \]

### Final Results
Thus, the kinetic energies of the alpha particle and the radon nucleus are approximately:

- \( K_{\alpha} \approx 4.79 \, \text{MeV} \)
- \( K_{Rn} \approx 0.08 \, \text{MeV} \)