Thorium-232 decays to lead-208 through a series of nuclear reactions. Only $\alpha$ particles and $\beta$ particles are emitted. How many $\beta$ particles are emitted?

Question

Quizplus · Accepted Answer

4Thorium-232 decays to lead-208 through a series of alpha (α) and beta (β) decays. Each alpha decay decreases the atomic number by 2 and the mass number by 4, while each beta decay increases the atomic number by 1 without changing the mass number. To go from thorium-232 (atomic number 90) to lead-208 (atomic number 82), six alpha decays occur (since 232 - 6*4 = 208), reducing the atomic number by 12 in total (2 per alpha decay). However, since the atomic number only needs to decrease by 8 (from 90 to 82), this means there must be four beta decays to increase the atomic number back up by 4 (90 - 12 + 4 = 82). Therefore, 4 beta particles are emitted.

Thorium-232 Decays to Lead-208 Through a Series of Nuclear Reactions α\alphaα

Thorium-232 Decays to Lead-208 Through a Series of Nuclear Reactions $\alpha$