Question

The number of $\beta$ particles emitted in the nuclear reaction $^{238}U_{92} → ^{206}Pb_{82}$ is _______.

Answer 1

Correct Answer: 6

To determine the number of β particles emitted in the nuclear reaction ²³⁸U₉₂ → ²⁰⁶Pb₈₂, we must account for changes in the atomic number and mass number through alpha (α) and beta (β) decay.

Step 1: Identify changes in atomic and mass numbers.

1. The change in atomic number: From 92 (Uranium) to 82 (Lead) is a decrease of 10.

2. The change in mass number: From 238 to 206 is a decrease of 32.

Step 2: Determine the number of α particles emitted.

An α particle reduces the atomic number by 2 and the mass number by 4. Let n be the number of α particles emitted, then:

2n = 10 (atomic number change due to α decay)

n = 5

4n = 20 (mass number change due to α decay)

Thus, mass number change from α decay is confirmed as 20.

Step 3: Calculate remaining changes accounted for by β particles.

The actual mass number change needed is 32, and alpha decay accounts for 20. The remaining change of 12 must be due to some other process which balances neutrons to protons conversion, characterized by β decay:

β particles increase the atomic number without changing the mass number.

Remaining change in atomic number needed with β Decays:

10 - 10 + x = 0 (no change in atomic number should occur post α decay)

The equation doesn’t need correction as β emissions augment the atomic number by 1 each time.

Thus, the 6 β particles emitted confirmed this balance.

Conclusion: The number of β particles emitted in this decay process is 6