In a nuclear reaction of `alpha` - decay, the daughter nuclei `._(Z)^(A)X` is moving with kinetic energy E. The total energy released if parent nuclei was at rest will be

Options

A. `E(1 + 4/A)`
B. `E/A (A)`
C. `E/4(A - 4)`
D. `E(A + 4)`

Answer

To solve the problem of determining the total energy released in an alpha decay nuclear reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Alpha Decay**: In alpha decay, a parent nucleus emits an alpha particle (which consists of 2 protons and 2 neutrons) and transforms into a daughter nucleus. The general reaction can be represented as:
\[ _{Z}^{A}X \rightarrow _{Z-2}^{A-4}Y + _{2}^{4}\alpha \]
Here, \(X\) is the parent nucleus, \(Y\) is the daughter nucleus, and \(\alpha\) is the emitted alpha particle.

2. **Conservation of Energy**: The total energy released in the decay process can be expressed in terms of the mass defect and the kinetic energy of the emitted particles. The energy released \(Q\) can be given by:
\[ Q = (m_{parent} - m_{daughter} - m_{\alpha})c^2 \]
where \(m_{parent}\), \(m_{daughter}\), and \(m_{\alpha}\) are the masses of the parent nucleus, daughter nucleus, and alpha particle respectively, and \(c\) is the speed of light.

3. **Kinetic Energy of Daughter Nucleus**: The daughter nucleus \(Y\) is moving with kinetic energy \(E\). According to conservation of momentum, the momentum of the emitted alpha particle and the daughter nucleus must be equal and opposite. Thus, if \(v_{\alpha}\) is the velocity of the alpha particle and \(v_{Y}\) is the velocity of the daughter nucleus, we can write:
\[ m_{Y}v_{Y} + m_{\alpha}v_{\alpha} = 0 \]
This implies:
\[ m_{Y}v_{Y} = -m_{\alpha}v_{\alpha} \]

4. **Relating Kinetic Energies**: The kinetic energy of the daughter nucleus can be expressed as:
\[ E = \frac{1}{2} m_{Y} v_{Y}^2 \]
and for the alpha particle:
\[ E_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]

5. **Total Energy Released**: The total energy released in the decay is the sum of the kinetic energies of the daughter nucleus and the alpha particle, along with the energy equivalent of the mass defect:
\[ Q = E + E_{\alpha} \]

6. **Final Expression**: Since we know that the kinetic energy of the daughter nucleus is \(E\), we can express the total energy released as:
\[ Q = E + \text{(kinetic energy of alpha particle)} \]

### Conclusion:
The total energy released in the alpha decay process, when the parent nucleus is at rest, is the sum of the kinetic energy of the daughter nucleus and the kinetic energy of the emitted alpha particle, along with the energy equivalent of the mass difference.