Number of α and β-particles emitted when  83222Bi is formed from  90238Th is :

Let mα− particle and nβ are - particles emitted. 90Th238→83Bi222+m2α4+n  −1β0On balancing number of neucleon,90=83+2m−n2m−n=7…… (1) 238=222+4m+n×04m = 16m = 4By equation (1)n=2m−7=8−7=1