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

Options

A. `4 alpha` and `7 beta`
B. `4alpha `and `1 beta`
C. `4 alpha` only
D. `7 beta `only

Answer

B

Explanation

`""_(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