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) rarr ._(Z - 6)B^(m - 12) + 3 ._(2)He^(4)`
Given weight of `A = m g`
`:.` Moles of `A (No) = 1 g` mol
Also `t = 20` days, `t_(1//2) = 10` days
`:.n = 2 (t = t_(1//2) xx n)`
`:.` Amount left in 2 halves `= (1)/(2^(2))"mol" = (1)/(4) "mol"`
`:.` Amount of decayed in 2 halves `= 1 - (1)/(4) = (3)/(4) "mol"`
`:.` Amount of `He` formed `= 3 xx (3)/(4) "mol" = (9)/(4) "mol"`
(decay of 1 mol gives 3 mol of `He`)
`:.` Volume of `He STP = (22.4 xx 9)/(4) = 50.4 L`