During the transformation of `._(c )X^(a)` to `._(d)Y^(b)` the number of `beta`-particles emitted are
a. `d + ((a - b)/(2)) - c` b. `(a - b)/(c )`
c. `d + ((a - b)/(2)) + c` d. `2c - d + a = b`

Options

A. a. `d + ((a - b)/(2)) - c`
B. b. `(a - b)/(c )`
C. c. `d + ((a - b)/(2)) + c`
D. d. `2c - d + a = b`

Explanation

To solve the problem, we need to analyze the nuclear transformation of the element \( _{c}X^{a} \) to \( _{d}Y^{b} \) and determine the number of beta particles emitted during this process.

### Step-by-Step Solution:

1. **Understanding the Transformation**:
- We have an initial nucleus \( _{c}X^{a} \) with atomic number \( c \) and mass number \( a \).
- It transforms into \( _{d}Y^{b} \) with atomic number \( d \) and mass number \( b \).
- During this transformation, beta particles (which are electrons or positrons) and possibly alpha particles (helium nuclei) may be emitted.

2. **Equating Atomic Numbers**:
- The atomic number of the initial nucleus is \( c \), and after the transformation, it becomes \( d \).
- The emission of beta particles increases the atomic number by 1 for each beta particle emitted. If \( m \) is the number of beta particles emitted, we can write:
\[ c - m + 2n = d \]
Here, \( n \) is the number of alpha particles emitted (each alpha particle decreases the atomic number by 2).

3. **Equating Mass Numbers**:
- The mass number of the initial nucleus is \( a \), and after transformation, it becomes \( b \).
- The mass number of beta particles is 0, and the mass number of alpha particles is 4. Thus, we can write:
\[ a = b + 4n \]

4. **Substituting for \( n \)**:
- From the mass number equation, we can express \( n \):
\[ n = \frac{a - b}{4} \]

5. **Substituting \( n \) into the Atomic Number Equation**:
- Substituting \( n \) back into the atomic number equation gives:
\[ c - m + 2\left(\frac{a - b}{4}\right) = d \]
- This simplifies to:
\[ c - m + \frac{a - b}{2} = d \]

6. **Rearranging the Equation**:
- Rearranging the equation to solve for \( m \):
\[ m = c - d + \frac{a - b}{2} \]

7. **Final Expression**:
- Therefore, the number of beta particles emitted is given by:
\[ m = d + \frac{a - b}{2} - c \]

### Conclusion:
The correct option for the number of beta particles emitted during the transformation is:
**a. \( d + \frac{(a - b)}{2} - c \)**