A radioisotope `._(Z)A^(m) (t_(1//2) = 10` days) decays to give `._(z - 6)B^(m - 12)` stable atom along with `alpha`-particles. If `m g` of A are taken and kept in a sealed tube, how much He will accumulate in 20 days at `STP`.

Explanation

`._(Z)A^(m) to ._(Z - 6)B^(m - 12) + 3 [._(2)He^(4)]`
Mole of `A = (m)/(m) = 1`
Number of half lives = 20/10 = 2
`N = N_(0) ((1)/(2))^(n) = 1 ((1)/(2))^(2) = (1)/(4)`
Decayed moles = 1 1/4 = 3/4
Mole of 'He' formed `= 3 xx 3//4 = 9//4`
volume of 'He' at STP `= 22.4 xx (9)/(4) = 50.4` litre