During alpha-decay , a nucleus decays by emitting an `alpha`-particle ( a helium nucleus `._2He^4`) according to the equation
`._Z^AX to ._(Z-2)^(A-4)Y+._2^4He+Q`
In this process, the energy released Q is shared by the emitted `alpha`-particle and daughter nucleus in the form of kinetic energy .
The energy Q is divided in a definite ratio among the `alpha`-particle and the daughter nucleus .
A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo `beta`-decay .This process also involves a release of definite energy . Initially, the `beta`-decay was represented as `._Z^AX to ._(Z+1)^AY + e^(-)"(electron)"+Q`
According to this reaction, the energy released during each decay must be divided in definite ratio by the emitted e' (`beta`-particle) and the daughter nucleus. While , in alpha decay, it has been found that every emitted `alpha`-particle has the same sharply defined kinetic energy. It is not so in case of `beta`-decay . The energy of emitted electrons or positrons is found to vary between zero to a certain maximum value. Wolfgang Pauli first suggested the existence of neutrinoes in 1930. He suggested that during `beta`-decay, a third particle is also emitted. It shares energy with the emitted `beta` particles and thus accounts for the energy distribution.
When a nucleus of mass number A at rest decays emitting an `alpha`-particle , the daugther nucleus recoils with energy K . What is the Q value of the reaction ?

Options

A. K
B. 2K
C. `"4K"/A`
D. `"AK"/4`

Answer

To solve the problem of finding the Q value of the reaction during alpha decay, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- During alpha decay, a nucleus with mass number \( A \) emits an alpha particle (which has a mass number of 4) and transforms into a daughter nucleus with mass number \( A - 4 \).
- The decay can be represented as:
\[ _Z^A X \rightarrow _{Z-2}^{A-4} Y + _2^4 He + Q \]

2. **Conservation of Momentum**:
- Initially, the parent nucleus is at rest, so its initial momentum is 0.
- After the decay, let the velocity of the alpha particle be \( v_1 \) and the velocity of the daughter nucleus be \( v_2 \).
- The masses are:
- Mass of the alpha particle = \( 4m \)
- Mass of the daughter nucleus = \( (A - 4)m \)
- By conservation of momentum:
\[ 4m \cdot v_1 = (A - 4)m \cdot v_2 \]
- Simplifying gives:
\[ 4v_1 = (A - 4)v_2 \quad \text{(Equation 1)} \]

3. **Kinetic Energy Calculation**:
- The kinetic energy of the daughter nucleus \( K \) is given as:
\[ K = \frac{1}{2}(A - 4)m v_2^2 \]
- The kinetic energy of the alpha particle is:
\[ KE_{\alpha} = \frac{1}{2}(4m)v_1^2 \]

4. **Substituting for \( v_1 \)**:
- From Equation 1, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{(A - 4)}{4} v_2 \]
- Substituting this into the kinetic energy expression for the alpha particle:
\[ KE_{\alpha} = \frac{1}{2}(4m)\left(\frac{(A - 4)}{4} v_2\right)^2 = \frac{1}{2}(4m) \cdot \frac{(A - 4)^2}{16} v_2^2 \]
\[ KE_{\alpha} = \frac{(A - 4)^2 m v_2^2}{8} \]

5. **Total Kinetic Energy**:
- The total kinetic energy released in the decay is:
\[ Q = K + KE_{\alpha} \]
- Substituting the expressions for \( K \) and \( KE_{\alpha} \):
\[ Q = \frac{1}{2}(A - 4)m v_2^2 + \frac{(A - 4)^2 m v_2^2}{8} \]

6. **Factoring Out Common Terms**:
- Factoring out \( m v_2^2 \):
\[ Q = m v_2^2 \left(\frac{1}{2}(A - 4) + \frac{(A - 4)^2}{8}\right) \]
- Simplifying the expression inside the parentheses:
\[ Q = m v_2^2 \left(\frac{4(A - 4) + (A - 4)^2}{8}\right) \]
- This can be simplified further, but for our purposes, we can relate it back to the kinetic energy \( K \) of the daughter nucleus.

7. **Final Expression for Q**:
- After some algebraic manipulation, we find:
\[ Q = \frac{A}{4} K \]

### Final Answer:
\[ Q = \frac{A}{4} K \]