Question

After the emission of a β-particle followed by an α-particle from \(^{214}_{83}\text{Bi}\), the number of neutrons in the atom is:

• A. 210
• B. 128
• C. 129
• D. 82

Answer 1

The Correct Option is B

The atomic number of \(^{214}_{83}\text{Bi}\) is 83, and the mass number is 214.

1. Emission of a β-particle: A β-particle emission increases the atomic number by 1 (83 → 84) but keeps the mass number unchanged (214).
2. Emission of an α-particle: An α-particle emission decreases the atomic number by 2 (84 → 82) and the mass number by 4 (214 → 210).

The number of neutrons is calculated as:

Number of neutrons = Mass number − Atomic number = 210 − 82 = 128

Answer 2

Correct Answer:

Option 2: 128

Explanation:

1. Beta Decay (β-particle emission):

²¹⁴₈₃Bi → ²¹⁴₈₄Po + β^- + ν̄_e

Atomic number increases by 1, mass number remains the same.

2. Alpha Decay (α-particle emission):

²¹⁴₈₄Po → ²¹⁰₈₂Pb + ⁴₂He

Atomic number decreases by 2, mass number decreases by 4.

3. Resulting Nucleus:

The resulting nucleus is ²¹⁰₈₂Pb (Lead).

4. Number of Neutrons:

Number of neutrons (N) = Mass number (A) - Atomic number (Z)

N = 210 - 82 = 128