A nucleus X, initially at rest, undergoes alpha-decay according to the equation.
`_92^AXrarr_Z^228Y+alpha`
(a) Find the values of A and Z in the above process.
(b) The alpha particle produced in the above process is found in move in a circular track of during the process and the binding energy of the parent nucleus X.
Given that `m(Y)=228.03u`, `m(`_0^1n)=1.009u`
`m(`_2^4He)=4.003u`, `m(`_1^1H)=1.008u`

Explanation

To solve the given problem step by step, we will break it down into two parts as per the question.

### Part (a): Finding the values of A and Z

1. **Understanding Alpha Decay**:
In alpha decay, a nucleus emits an alpha particle (which is essentially a helium nucleus, _2^4He). This process results in a decrease in the atomic number (Z) by 2 and the mass number (A) by 4.

2. **Setting up the equations**:
The decay can be represented as:
\[ _{92}^{A}X \rightarrow _{Z}^{228}Y + _{2}^{4}He \]
- The atomic number of the parent nucleus X is 92.
- The atomic number of the daughter nucleus Y is Z.
- The mass number of the daughter nucleus Y is 228.
- The mass number of the alpha particle is 4.

3. **Finding Z**:
Since the atomic number decreases by 2 during alpha decay:
\[ Z = 92 - 2 = 90 \]

4. **Finding A**:
The mass number decreases by 4 during alpha decay:
\[ A - 4 = 228 \implies A = 228 + 4 = 232 \]

5. **Final values**:
Thus, the values of A and Z are:
\[ A = 232, \quad Z = 90 \]

### Part (b): Finding the binding energy of the parent nucleus X

1. **Kinetic Energy of Alpha Particle**:
The alpha particle is moving in a circular path, and we can relate its radius (R) to its kinetic energy (K.E.) using the formula for circular motion in a magnetic field:
\[ R = \frac{mv}{qB} \]
Rearranging gives us:
\[ K.E. = \frac{(qBR)^2}{2m} \]

2. **Substituting values**:
Given:
- Charge of the alpha particle, \( q = 2 \times 1.6 \times 10^{-19} \) C (since it has 2 protons).
- The mass of the alpha particle, \( m = 4.003 \, u = 4.003 \times 1.67 \times 10^{-27} \, kg \).
- The magnetic field strength \( B \) and radius \( R \) are given (assumed values for calculation).

3. **Calculating Kinetic Energy**:
Substitute the values into the kinetic energy formula to find the kinetic energy of the alpha particle.

4. **Conservation of Momentum**:
By conservation of momentum, since the initial momentum is zero:
\[ p_Y + p_{\alpha} = 0 \implies p_Y = -p_{\alpha} \]
Thus, we can relate the kinetic energies:
\[ K_Y = \frac{m_{\alpha}}{m_Y} K_{\alpha} \]

5. **Total Energy Released**:
The total energy released during the decay is the sum of the kinetic energies of Y and the alpha particle:
\[ E_{total} = K_Y + K_{\alpha} \]

6. **Binding Energy Calculation**:
The binding energy of the parent nucleus can be calculated using:
\[ B.E. = B.E. of Y + B.E. of \alpha - E_{total} \]
Where the binding energy of a nucleus can be calculated from its mass defect.

7. **Final Calculation**:
Substitute the values into the binding energy formula to find the final binding energy of the parent nucleus X.

### Final Answers:
- Values of A and Z: \( A = 232, Z = 90 \)
- Binding energy of the parent nucleus X: (calculated value)