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 determine the atomic number (Z) of the resulting nucleus after a series of emissions, we will analyze each type of emission step by step.

### Step 1: Initial Atomic Number
The initial atomic number (Z) of the nucleus is given as 92.

### Step 2: Emission of Alpha Particles
Alpha particles (α) are emitted first. Each alpha particle decreases the atomic number by 2.

- **Number of alpha particles emitted**: 8
- **Change in atomic number due to alpha emissions**:
\[ \text{Change} = 8 \times (-2) = -16 \]

### Step 3: Emission of Beta Minus Particles
Beta minus particles (β⁻) are emitted next. Each beta minus emission increases the atomic number by 1.

- **Number of beta minus particles emitted**: 4
- **Change in atomic number due to beta minus emissions**:
\[ \text{Change} = 4 \times (+1) = +4 \]

### Step 4: Emission of Beta Plus Particles
Beta plus particles (β⁺) are emitted last. Each beta plus emission decreases the atomic number by 1.

- **Number of beta plus particles emitted**: 2
- **Change in atomic number due to beta plus emissions**:
\[ \text{Change} = 2 \times (-1) = -2 \]

### Step 5: Calculate the Final Atomic Number
Now, we can calculate the final atomic number by combining all the changes:

1. Start with the initial atomic number:
\[ Z_{\text{initial}} = 92 \]

2. Apply the changes from alpha emissions:
\[ Z = 92 - 16 = 76 \]

3. Apply the changes from beta minus emissions:
\[ Z = 76 + 4 = 80 \]

4. Apply the changes from beta plus emissions:
\[ Z = 80 - 2 = 78 \]

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

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