How many alpha and beta particles are emitted when Uranium ₉₂U²³⁸ decays to lead ₈₂Pb²⁰⁶ ?

Solution

Physics sir · 4 steps

Beta decay never changes the mass number, so fix the alphas from A first, then balance Z with betas.

1. What each decay does. An alpha particle (⁴₂He) lowers A by 4 and Z by 2. A beta-minus particle leaves A unchanged and raises Z by 1, because a neutron turns into a proton.
2. Count the alphas. Only alpha decay changes the mass number. A falls from 238 to 206, a drop of 32, so the number of alphas is 32 ÷ 4 = 8.

   Z_(final) = 92 - 2(8) + n_β = 82

3. Count the betas. Eight alphas alone take Z down to 92 - 16 = 76. Lead has Z = 82, so 6 beta decays are needed to raise Z by 6.
4. So. The decay emits 8 alpha particles and 6 beta particles, which is option (d).

The idea to keep. In a decay chain, the change in mass number counts the alphas, since betas do not touch A. Then use the atomic number to find how many betas balance the charge.

Answer from the NTA final answer key. Solution by PrepWiser.