If a nucleus such as `.^(226)Ra` that is initially at rest undergoes `alpha`- decay , then which of the following statemnets is true?

Options

A. The alpha particles has more kinetic than the daughter nuclues.
B. The alpha particle has less kinetic energy than the daughter nucleus.
C. The alpha particle and daughter nucleus both have same kinetic energy
D. We cannot say anything about kinetic energy of alpha particle and daughter nucleus.

Explanation

To solve the question about the alpha decay of a nucleus such as \( ^{226}\text{Ra} \), we will analyze the situation using the principles of conservation of momentum and kinetic energy.

### Step-by-Step Solution:

1. **Understanding Alpha Decay**:
- In alpha decay, a nucleus emits an alpha particle (which consists of 2 protons and 2 neutrons) and transforms into a daughter nucleus. For \( ^{226}\text{Ra} \), the decay can be represented as:
\[ ^{226}\text{Ra} \rightarrow ^{222}\text{Rn} + \alpha \]
- Here, \( ^{222}\text{Rn} \) is the daughter nucleus and \( \alpha \) is the emitted alpha particle.

2. **Initial Conditions**:
- The nucleus \( ^{226}\text{Ra} \) is initially at rest, which means its initial momentum is zero.

3. **Conservation of Momentum**:
- According to the law of conservation of momentum, the total momentum before decay must equal the total momentum after decay. Since the initial momentum is zero, we have:
\[ 0 = p_{\alpha} + p_{\text{daughter}} \]
- This implies:
\[ p_{\alpha} = -p_{\text{daughter}} \]
- Therefore, the magnitudes of the momenta are equal:
\[ |p_{\alpha}| = |p_{\text{daughter}}| \]

4. **Relating Kinetic Energy to Momentum**:
- The kinetic energy (KE) of an object can be expressed in terms of its momentum (p) and mass (m):
\[ KE = \frac{p^2}{2m} \]
- For the alpha particle and the daughter nucleus, we can write:
\[ KE_{\alpha} = \frac{p_{\alpha}^2}{2m_{\alpha}} \quad \text{and} \quad KE_{\text{daughter}} = \frac{p_{\text{daughter}}^2}{2m_{\text{daughter}}} \]

5. **Comparing Kinetic Energies**:
- Since \( |p_{\alpha}| = |p_{\text{daughter}}| \), we can substitute \( p_{\text{daughter}} \) with \( p_{\alpha} \):
\[ KE_{\alpha} = \frac{p_{\alpha}^2}{2m_{\alpha}} \quad \text{and} \quad KE_{\text{daughter}} = \frac{p_{\alpha}^2}{2m_{\text{daughter}}} \]
- Thus, we can compare the two kinetic energies:
\[ \frac{KE_{\alpha}}{KE_{\text{daughter}}} = \frac{m_{\text{daughter}}}{m_{\alpha}} \]
- Since the mass of the daughter nucleus (\( m_{\text{daughter}} \)) is greater than the mass of the alpha particle (\( m_{\alpha} \)), it follows that:
\[ KE_{\alpha} > KE_{\text{daughter}} \]

6. **Conclusion**:
- From the analysis, we conclude that the alpha particle has more kinetic energy than the daughter nucleus. Therefore, the correct statement is:
- **A**: The alpha particle has more kinetic energy than the daughter nucleus.