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

Options

A. 80
B. 81
C. 82
D. 83

Explanation

To find the atomic number of the resulting nucleus after the emission of the specified particles, we will analyze the effect of each particle emitted on the atomic number (Z) of the nucleus.

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

2. **Emission of Particles**: We will go through each particle emitted in the sequence and adjust the atomic number accordingly:
- **Alpha Particle (α)**: Emits 2 protons and 2 neutrons. This decreases the atomic number by 2.
- **Beta Minus Particle (β⁻)**: This is an electron emitted from the nucleus, which increases the atomic number by 1.
- **Beta Plus Particle (β⁺)**: This is a positron emitted from the nucleus, which decreases the atomic number by 1.

3. **Sequence of Emissions**:
- **1st emission: α**: Z = 92 - 2 = 90
- **2nd emission: β⁻**: Z = 90 + 1 = 91
- **3rd emission: α**: Z = 91 - 2 = 89
- **4th emission: α**: Z = 89 - 2 = 87
- **5th emission: β⁻**: Z = 87 + 1 = 88
- **6th emission: β⁺**: Z = 88 - 1 = 87
- **7th emission: α**: Z = 87 - 2 = 85
- **8th emission: β⁻**: Z = 85 + 1 = 86
- **9th emission: α**: Z = 86 - 2 = 84
- **10th emission: β⁺**: Z = 84 - 1 = 83

4. **Final Atomic Number**: After all emissions, the final atomic number of the resulting nucleus is 83.

Thus, the atomic number of the resulting nucleus is **83**.