Calculate the total number of α- and β-particles emitted in the nuclear reaction 92U238⟶82Pb214.
(IIT-JEE, 2009)
92U238→ 82Pb214+x 2He4+y −1e0
Equating atomic number on both sides
238 = 214 + x x 4 + y x 0
∴ x = 6
Hence, 6 α-particles are emitted.
Equating atomic number on both sides
92 = 82 + 6 x 2 + y (-1)
∴ y = 2
Hence, 2 β-particles are emitted.
Therefore, total 8 particles (6α, 2β) are emitted.