A nucleus with Z =92 emits the following in a sequence:
`alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha`. The Z of the resulting nucleus is

Options

A. 74
B. 76
C. 78
D. 82

Explanation

To find the atomic number \( Z \) of the resulting nucleus after a series of decays, we need to analyze the effects of each type of decay on the atomic number. The initial atomic number is given as \( Z = 92 \).

### Step-by-Step Solution:

1. **Identify the Decay Types**:
- **Alpha Decay**: Reduces the atomic number \( Z \) by 2.
- **Beta Minus Decay (\( \beta^- \))**: Increases the atomic number \( Z \) by 1.
- **Beta Plus Decay (\( \beta^+ \))**: Decreases the atomic number \( Z \) by 1.

2. **Count the Number of Each Decay**:
- **Alpha Decays**: There are 8 alpha decays.
- **Beta Minus Decays**: There are 4 beta minus decays.
- **Beta Plus Decays**: There are 2 beta plus decays.

3. **Calculate the Change in Atomic Number**:
- For **Alpha Decays**:
\[ \text{Change from alpha decays} = -2 \times 8 = -16 \]
- For **Beta Minus Decays**:
\[ \text{Change from beta minus decays} = +1 \times 4 = +4 \]
- For **Beta Plus Decays**:
\[ \text{Change from beta plus decays} = -1 \times 2 = -2 \]

4. **Total Change in Atomic Number**:
\[ \text{Total change} = -16 + 4 - 2 = -14 \]

5. **Calculate the Resulting Atomic Number**:
\[ Z_{\text{final}} = Z_{\text{initial}} + \text{Total change} = 92 - 14 = 78 \]

### Final Result:
The atomic number \( Z \) of the resulting nucleus is **78**.

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