`._(84)Po^(210)` decays with `alpha-` particle to `._(82)Pb^(206)` with a half-line of `138.4` day. If `1.0g` of `._(84)Po^(210)` is placed in a sealed tube, how much helium will accumulate in `69.2` day? Express the answer in `cm^(3)` at `STP`. Also report the volume of `He` formed if `1g` of `Po^(210) O_(2)` is used.

Explanation

`t_(1//2) = 138.4` day, `t = 69.2` day
`:.` No. of halves `n = (t)/(t_(1//2)) = (69.2)/(138.4) = (1)/(2)`
`:.` Amount of `Po` left after `(1)/(2)` halves `= (1)/((2)^(1//2))g = 0.707g`
`:.` Amount of `Po` used in `(1)/(2)` haves `= 1 - 0.707 = 0.293g`
Now `._(84)Po^(210) rarr ``._(82)Pb^(206) + ``_(2)He^(4)`
`:' 210 g Po` on decay will produce `= 4g He`
`:. 0.293 g Po` on decay will produce
`= (4xx0.293)/(210) = 5.581xx10^(3)g He`
`:.` Volume of `He` at `STP = (5.581xx10^(-3)xx22400)/(4)`
`= 31.25 mL = 31.25 cm^(3)`
Also Amount of `Po^(210)` in `1g PoO_(2) = (210)/(242) = 0.868`
`:.` Amount of `Po^(210)` left after `1//2` halves
`= [(210)/(242)] xx (1)/(2^(1//2)) = 0.614g`
`:.` Amount of `Po^(210)` used after `1//2` halves
`= 0.868-0.614 = 0.254g`
`:.` Amount of `He` formed `= (4xx0.254)/(210) = 4.84xx10^(-3)g`
`:.` Volume of `He` at `STP`
`= (4.84xx10^(-3)xx22400)/(4) = 27.104 cm^(3)`