The number of `beta`-particles emitted during the change `"_aX^c` `rarr "_dY^b+m_2He^4+n_(-1) e^0` is

Options

A. `(a-b)/4`
B. `d+[(c-b)/2]-a`
C. `d+[(a-b)/2]-a`
D. `d+[(a-b)/2]-c`

Answer

To solve the problem of determining the number of beta particles emitted during the nuclear reaction \( _aX^c \rightarrow _dY^b + m_2He^4 + n_{-1}e^0 \), we will follow these steps:

### Step 1: Identify the components of the reaction
The reaction involves:
- A parent nucleus \( _aX^c \)
- A daughter nucleus \( _dY^b \)
- An alpha particle \( m_2He^4 \)
- Beta particles \( n_{-1}e^0 \)

### Step 2: Apply conservation of mass number
The mass number must be conserved in the reaction. This means:
\[ c = b + 4m + 0n \]
Since \( n \) (the number of beta particles) does not contribute to the mass number, we can simplify this to:
\[ c = b + 4m \]

### Step 3: Apply conservation of atomic number
The atomic number must also be conserved. This gives us the equation:
\[ a = d + 2m - n \]
Here, \( 2m \) accounts for the two protons in the alpha particle emitted.

### Step 4: Rearranging the equations
From the mass number conservation, we can express \( m \):
\[ m = \frac{c - b}{4} \]
Now, substitute this value of \( m \) into the atomic number conservation equation:
\[ a = d + 2\left(\frac{c - b}{4}\right) - n \]
This simplifies to:
\[ a = d + \frac{c - b}{2} - n \]

### Step 5: Solve for \( n \)
Rearranging the equation to solve for \( n \):
\[ n = d + \frac{c - b}{2} - a \]

### Step 6: Final expression for the number of beta particles
Thus, the number of beta particles emitted, \( n \), can be expressed as:
\[ n = d + \frac{c - b}{2} - a \]

### Conclusion
The final expression gives us the number of beta particles emitted during the reaction.