`""_86^222Rn` goes through radioactive distintegrations by successive emissions of `alpha, alpha, beta, beta , alpha , beta " and " beta` particles. Then final nucleus is

Options

A. a. `""_82^206 Pb`
B. b. `""_84^210Po`
C. c.`""_86^211Ra`
D. d.`""_83^209Bi`

Answer

B

Explanation

For `R_n , A = 222 " and " Z = 86`
it emits `3alpha` particles and `4beta` particles
`therefore` Its new atomic number will be
Z' = 86 - `3xx2+4xx1=84`
and its new mass number A' = 222 - `3xx4 = 210`
`therefore` The final nucleus is `""_84^210Po`