Question

`""_(84)Po""^(210)` decays with a particle to `""_(82)Pb""^(206)` with a half life of 138.4 days. If 1.0 g of `""_(84)Po""^(210)` is placed in a sealed tube, how much helium will accumulate in 69.2 days. Express the answer in `cm""^(3)` at STP

Answer 1

`t""_(1//2)` = 138.4 days , t = 69.2 day
`therefore` Number of half - lifes n = `(t)/t""_(1//2) = (69.2)/(138.4)=(1)/(2)`
`therefore` Amount of Po left after `(1)/(2)` half life
`=(1)/((2)""^(1//2))` g = 0.707g
`therefore` Amount of Po used in `(1)/(2)` half life
= 1-0.707 = 0.293 g
Now `""_(84)Po""^(210)to .""_(82)Pb""^(206)+""_(2)He""^(4)`
`because` 210 g Po in decay will produce = 4g He
`therefore` 0.293 g Po on decay will produce =
`(4xx0.293)/210 = 5.581 xx 10""^(-3)g` He
`therefore` Volume of He at STP =
`(5.581xx10""^(-3)xx22400)/(4)=31.25mL =31.25cm""^(3)`