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.
Which of the following is correct ?

Options

A. `m_(x) = m_(y) + m _(alpha)`
B. `m_(x) gt m_(y) + m _(alpha)`
C. `m_(x) lt m_(y) + m _(alpha)`
D. none of the above

Answer

To solve the problem, we need to analyze the decay of a radioactive nucleus and understand the relationship between the masses of the parent nucleus and the decay products, as well as the kinetic energy produced in the process.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The decay of the radioactive nucleus can be represented as:
\[ x \rightarrow y + \alpha \]
where \(x\) is the parent nucleus, \(y\) is the daughter nucleus, and \(\alpha\) is the emitted alpha particle.

2. **Conservation of Energy**:
Initially, the nucleus \(x\) is at rest, which means its initial kinetic energy is zero. After the decay, the daughter nucleus \(y\) and the alpha particle \(\alpha\) will have kinetic energy. The energy released during the decay must come from the difference in mass between the parent nucleus and the decay products.

3. **Mass-Energy Equivalence**:
According to Einstein's mass-energy equivalence principle, the energy \(E\) released during the decay can be expressed as:
\[ E = \Delta m \cdot c^2 \]
where \(\Delta m\) is the change in mass, defined as:
\[ \Delta m = m_x - (m_y + m_\alpha) \]
Here, \(m_x\) is the mass of the parent nucleus, \(m_y\) is the mass of the daughter nucleus, and \(m_\alpha\) is the mass of the alpha particle.

4. **Condition for Energy Release**:
For energy to be released during the decay, the change in mass \(\Delta m\) must be positive:
\[ \Delta m > 0 \implies m_x > m_y + m_\alpha \]
This means that the mass of the parent nucleus must be greater than the total mass of the daughter nucleus and the alpha particle combined.

5. **Kinetic Energy Distribution**:
The energy \(E\) released is converted into the kinetic energy of the daughter nucleus \(y\) and the alpha particle \(\alpha\). Therefore, the kinetic energy of the alpha particle just after the decay, denoted as \(E_0\), is a portion of this released energy.

6. **Conclusion**:
Based on the above analysis, we conclude that the correct relationship between the masses is:
\[ m_x > m_y + m_\alpha \]
This means that the correct answer is option B.