Heavy radioactive nucleus decay through `alpha-`decay also. Consider a radioactive nucleus x. It spontaneously undergoes decay at rest resulting in the formation of a daughter nucleus y and the emission of an `alpha-` particle. The radioactive reaction can be given by
`x rarr y + alpha`
However , the nucleus y and the `alpha-` particle will be in motion. Then a natural question arises that what provides kinetic energy to the radioactive products. In fact, the difference of masses of the decaying nucleus and the decay products provides for the energy that is shared by the daughter nucleus and the `alpha-` particle as kinetic energy . We know that Einstein's mass-energy equivalence relation `E = m c^(2)`. Let `m_(x), m_(y)` and `m_(alpha)` be the masses of the parent nucleus x, the daughter nucleus y and `alpha-` particle respectively. Also the kinetic energy of `alpha-` particle just after the decay is `E_(0)`. Assuming all motion of to be non-relativistic.
Just after the decy , if the speed of `alpha-` particle is v, then the speed of the centre of mass of the system of the daughter nucleud y and the `alpha-` particle will be

Options

A. v
B. `( m_(alpha))/( m_(y) + m_(y)) v `
C. `(m_(y))/( m_(y) + m_(x)) v `
D. none of the above

Answer

To find the speed of the center of mass of the system consisting of the daughter nucleus \( Y \) and the emitted alpha particle after the decay of the parent nucleus \( X \), we can follow these steps:

### Step 1: Understand the Initial Conditions
Initially, the parent nucleus \( X \) is at rest. Therefore, its initial momentum is zero.

### Step 2: Apply Conservation of Momentum
According to the law of conservation of momentum, the total momentum before the decay must equal the total momentum after the decay. Since the parent nucleus is at rest, we have:
\[ m_x \cdot 0 = m_y \cdot v_y + m_{\alpha} \cdot v_{\alpha} \]
Where:
- \( m_x \) is the mass of the parent nucleus \( X \)
- \( m_y \) is the mass of the daughter nucleus \( Y \)
- \( m_{\alpha} \) is the mass of the alpha particle
- \( v_y \) is the velocity of the daughter nucleus \( Y \)
- \( v_{\alpha} \) is the velocity of the alpha particle

### Step 3: Rearranging the Momentum Equation
From the momentum conservation equation, we can express the velocity of the daughter nucleus \( Y \):
\[ 0 = m_y \cdot v_y + m_{\alpha} \cdot v_{\alpha} \]
This implies:
\[ m_y \cdot v_y = -m_{\alpha} \cdot v_{\alpha} \]
Thus,
\[ v_y = -\frac{m_{\alpha}}{m_y} \cdot v_{\alpha} \]

### Step 4: Calculate the Velocity of the Center of Mass
The velocity of the center of mass \( v_{cm} \) of the system can be calculated using the formula:
\[ v_{cm} = \frac{m_y \cdot v_y + m_{\alpha} \cdot v_{\alpha}}{m_y + m_{\alpha}} \]

### Step 5: Substitute \( v_y \) into the Center of Mass Equation
Substituting \( v_y \) from step 3 into the center of mass velocity equation:
\[ v_{cm} = \frac{m_y \left(-\frac{m_{\alpha}}{m_y} \cdot v_{\alpha}\right) + m_{\alpha} \cdot v_{\alpha}}{m_y + m_{\alpha}} \]
This simplifies to:
\[ v_{cm} = \frac{-m_{\alpha} \cdot v_{\alpha} + m_{\alpha} \cdot v_{\alpha}}{m_y + m_{\alpha}} = \frac{0}{m_y + m_{\alpha}} = 0 \]

### Conclusion
The speed of the center of mass of the system of the daughter nucleus \( Y \) and the alpha particle is:
\[ \boxed{0} \]