A stationary radioactive nucleus of mass 210 units disintegrates into an alpha particle of mass 4 units and residual nucleus of mass 206 units. If the kinetic energy of the alpha particle is E, then the kinetic energy of the residual nucleus is

Options

A. `((2)/(105))E`
B. `((2)/(103))E`
C. `((103)/(105))E`
D. `((103)/(2))E`

Explanation

To solve the problem, we will use the principles of conservation of momentum and the relationship between kinetic energy and momentum.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a stationary radioactive nucleus that disintegrates into an alpha particle and a residual nucleus. The masses are given as follows:
- Mass of the original nucleus (M) = 210 units
- Mass of the alpha particle (m₁) = 4 units
- Mass of the residual nucleus (m₂) = 206 units
- Kinetic energy of the alpha particle (K₁) = E
- We need to find the kinetic energy of the residual nucleus (K₂).

2. **Conservation of Momentum**:
Since the original nucleus is stationary, the total momentum before disintegration is zero. Therefore, the total momentum after disintegration must also be zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]
where \(v_1\) is the velocity of the alpha particle and \(v_2\) is the velocity of the residual nucleus.

3. **Expressing Velocities**:
From the momentum equation, we can express the velocity of the residual nucleus in terms of the velocity of the alpha particle:
\[ m_1 v_1 = -m_2 v_2 \implies v_2 = -\frac{m_1}{m_2} v_1 \]

4. **Kinetic Energy Relation**:
The kinetic energy (K) of an object is given by:
\[ K = \frac{1}{2} mv^2 \]
For the alpha particle:
\[ K_1 = \frac{1}{2} m_1 v_1^2 = E \]
For the residual nucleus:
\[ K_2 = \frac{1}{2} m_2 v_2^2 \]

5. **Substituting for \(v_2\)**:
Substitute \(v_2\) into the kinetic energy expression for the residual nucleus:
\[ K_2 = \frac{1}{2} m_2 \left(-\frac{m_1}{m_2} v_1\right)^2 = \frac{1}{2} m_2 \frac{m_1^2}{m_2^2} v_1^2 \]
Simplifying gives:
\[ K_2 = \frac{1}{2} \frac{m_1^2}{m_2} v_1^2 \]

6. **Relating \(K_2\) to \(E\)**:
Since \(K_1 = \frac{1}{2} m_1 v_1^2 = E\), we can express \(v_1^2\) in terms of \(E\):
\[ v_1^2 = \frac{2E}{m_1} \]
Substitute this into the equation for \(K_2\):
\[ K_2 = \frac{1}{2} \frac{m_1^2}{m_2} \cdot \frac{2E}{m_1} = \frac{m_1 E}{m_2} \]

7. **Final Calculation**:
Plugging in the values:
- \(m_1 = 4\) units
- \(m_2 = 206\) units
\[ K_2 = \frac{4E}{206} = \frac{2E}{103} \]

### Conclusion:
The kinetic energy of the residual nucleus is:
\[ K_2 = \frac{2E}{103} \]