Q: How many α and β – particles are emitted when nucleus _{92} U^{238} decay to _{82} Pb^{214} ?

Sol: Let n be the number of – particle and m be the number of – particles emitted.

_{92} U^{238} _{82} Pb^{214} + n _{2} He^{4} + m _{-1} e^{0}

As mass is conserved, 238 = 214 + 4 n + m (0)

= 214 + 4n ; 4 n = 24 ; n = 6

As charge is conserved, 92 = 82 + 2n + m (-1)

10 = 2 (6) –m (∵ n = 6) ; m = 2

6 α – particles and 2 β – particles are emitted.