If ""_(90)Th^(238) disintegrates to ""_(83)Bi^(212) , then the number alpha and beta particles emitted is

If ""_(90)Th^(238) disintegrates to ""_(83)Bi^(212) , then the number

1.	If ""_(90)Th^(238) disintegrates to ""_(83)Bi^(212) , then the number alpha and beta particles emitted is
Answer» `4 alpha` and `7 BETA` `4alpha `and `1 beta` `4 alpha` only `7 beta `only Solution :`""_(90)Th^(228)rarr ""_(83)Bi^(211) +n_(2)He^(4)+m""_(-1)e^(0)` `THEREFORE 228 =212 +4N impliesn=4` 90=83 + 2n-m `impliesm=1` `therefore alpha`-particles emitted, n=4 `beta` particles emitted , m=1