Question

The sum of the number of \(\alpha\)-particles and \(\beta\)-particles emitted in the nuclear decay process, \(^{238}_{92}\mathrm{U} \rightarrow\, ^{206}_{82}\mathrm{Pb}\), is (in integer).

Hint:

Alpha decay drops mass number by 4 and atomic number by 2; beta decay keeps mass number fixed but raises atomic number by 1. Balance both numbers separately, then add the two particle counts.

Answer 1

Correct Answer: 14

A nuclear decay chain must conserve both mass number (top number) and atomic number (bottom number) at every step, so instead of writing individual equations, tabulate the total change needed and split it between the two particle types.

1. Total mass number drop: $238 - 206 = 32$. Only $\alpha$-emission changes mass number, and each $\alpha$ carries away 4 mass units. So the number of $\alpha$-particles is $32/4 = 8$.
2. Total atomic number drop: $92 - 82 = 10$. But $\alpha$-emission alone would drop $Z$ by $2 \times 8 = 16$, which overshoots the required drop of 10 by $16-10=6$.
3. Beta correction: every $\beta^-$ emitted pushes $Z$ back up by 1 (a neutron becomes a proton plus an ejected electron), so exactly 6 $\beta$-particles are needed to cancel that overshoot and land exactly on $Z=82$.

So the decay uses 8 $\alpha$-particles and 6 $\beta$-particles.

Let's summarize:

• Mass number balance alone fixes the alpha count: $(238-206)/4 = 8$.
• Atomic number balance then fixes the beta count once alpha's contribution is subtracted: $2(8) - (92-82) = 6$.

Adding them, $8+6=14$, so the total number of particles emitted is $\boxed{14}$.