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

Step 1: Set up the mass number and atomic number balance.
In the decay chain \(^{238}_{92}\mathrm{U} \rightarrow\, ^{206}_{82}\mathrm{Pb}\), let the number of \(\alpha\)-particles emitted be \(a\) and the number of \(\beta\)-particles (\(\beta^-\)) be \(b\). An \(\alpha\)-particle is a \(^4_2\mathrm{He}\) nucleus, so it removes 4 from the mass number and 2 from the atomic number. A \(\beta^-\)-particle is an electron ejected when a neutron converts to a proton, so it leaves the mass number unchanged but raises the atomic number by 1.

Step 2: Balance the mass number.
Only \(\alpha\)-decay changes the mass number.
\[ 238 - 4a = 206 \] \[ 4a = 32 \implies a = 8 \] So 8 \(\alpha\)-particles are emitted.

Step 3: Balance the atomic number.
Each \(\alpha\)-particle lowers Z by 2, and each \(\beta\)-particle raises Z by 1.
\[ 92 - 2a + b = 82 \] Put \(a = 8\):
\[ 92 - 16 + b = 82 \] \[ 76 + b = 82 \implies b = 6 \] So 6 \(\beta\)-particles are emitted.

Step 4: Add the two counts.
\[ a + b = 8 + 6 = 14 \]
Final Answer:
The sum of \(\alpha\) and \(\beta\)-particles emitted is \[ \boxed{14} \]