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

Options

A. 76
B. 79
C. 82
D. 74

Answer

To solve the problem, we need to track the changes in the atomic number (Z) of the nucleus as it emits particles in the given sequence. The initial atomic number is Z = 92. We will analyze the effect of each type of emission on Z.

### Step-by-Step Solution:

1. **Initial Atomic Number (Z)**:
- The initial atomic number is Z = 92.

2. **Alpha Emission**:
- Each alpha particle emission decreases Z by 2.
- There are 8 alpha emissions in total.
- Change in Z due to alpha emissions: \( 8 \times (-2) = -16 \).
- New Z after alpha emissions:
\[ Z = 92 - 16 = 76 \]

3. **Beta Minus Emission**:
- Each beta minus emission increases Z by 1 (as a neutron converts to a proton).
- There are 4 beta minus emissions.
- Change in Z due to beta minus emissions: \( 4 \times (+1) = +4 \).
- New Z after beta minus emissions:
\[ Z = 76 + 4 = 80 \]

4. **Beta Plus Emission**:
- Each beta plus emission decreases Z by 1 (as a proton converts to a neutron).
- There is 1 beta plus emission.
- Change in Z due to beta plus emissions: \( 1 \times (-1) = -1 \).
- New Z after beta plus emissions:
\[ Z = 80 - 1 = 79 \]

5. **Final Atomic Number (Z)**:
- After all emissions, the final atomic number is Z = 79.

### Final Answer:
The Z of the resulting nucleus is **79**.

---