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 total energy released in an alpha decay nuclear reaction, we will follow these steps:

### Step 1: Understand the decay process
In alpha decay, a parent nucleus \( _{Z}^{A}X \) emits an alpha particle (which is a helium nucleus, \( _{2}^{4}He \)) and transforms into a daughter nucleus \( _{Z-2}^{A-4}Y \). The initial momentum of the parent nucleus is zero since it is at rest.

### Step 2: Set up the conservation of momentum
Since the parent nucleus is at rest, the total initial momentum is zero. After the decay, the momentum of the daughter nucleus and the alpha particle must be equal in magnitude and opposite in direction:
\[ P_{\text{daughter}} + P_{\text{alpha}} = 0 \]
This implies:
\[ P_{\text{daughter}} = -P_{\text{alpha}} \]

### Step 3: Express momentum in terms of kinetic energy
The momentum \( P \) of an object can be related to its kinetic energy \( E \) using the formula:
\[ P = \sqrt{2mE} \]
where \( m \) is the mass of the object and \( E \) is its kinetic energy.

### Step 4: Write expressions for the momenta
Let \( m_d \) be the mass of the daughter nucleus and \( m_{\alpha} \) be the mass of the alpha particle (approximately 4 units). The kinetic energy of the daughter nucleus is given as \( E \).

For the daughter nucleus:
\[ P_{\text{daughter}} = \sqrt{2 m_d E} \]
For the alpha particle:
\[ P_{\text{alpha}} = \sqrt{2 \cdot 4 \cdot E} = \sqrt{8E} \]

### Step 5: Set the magnitudes of the momenta equal
From the conservation of momentum:
\[ \sqrt{2 m_d E} = \sqrt{8E} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 2 m_d E = 8E \]
Assuming \( E \neq 0 \), we can divide both sides by \( E \):
\[ 2 m_d = 8 \]
Thus, we find:
\[ m_d = 4 \]

### Step 7: Calculate the total energy released
The total energy released in the decay can be expressed as the sum of the kinetic energies of the daughter nucleus and the alpha particle. The kinetic energy of the daughter nucleus is \( E \), and the kinetic energy of the alpha particle can be expressed as:
\[ E_{\alpha} = \frac{1}{2} \cdot m_{\alpha} v_{\alpha}^2 \]
Using the relationship we derived earlier, we can express the total energy released as:
\[ E_{\text{total}} = E + E_{\alpha} \]
Substituting \( E_{\alpha} \) in terms of \( E \):
\[ E_{\text{total}} = E + 4E/A \]
Thus:
\[ E_{\text{total}} = E \left(1 + \frac{4}{A}\right) \]

### Final Answer
The total energy released in the alpha decay when the parent nucleus is at rest is:
\[ E_{\text{total}} = E \left(1 + \frac{4}{A}\right) \]