10 g-atoms of an `alpha`-active radioisotope are disintegrating in a sealed container. In one hour the helium gas collected at STP is `11.2 cm^(3)`. Calculate the half life of the radioisotope.

Explanation

Amount of radioactive isotope = 10g atoms
`N_(0) = 10 xx 6.023 xx 10^(23)` atoms
`= 6.023 xx 10^(24` atoms
22400 `cm^(3)` of helium contains `= 6.023 xx 10^(23)` atoms
`11.2 cm^(3)` of helium will contain `= (6.023 xx 10^(23))/(22400) xx 11.2` atoms
`= 3.01 xx 10^(20)` atoms
As one helium atom is obtained by disintegration of one atom of radioisotope, the total number of atoms of the radioactive isotope which have disintegrated in one hour
`= 3.01 xx 10^(20)` or `0.0003 xx 10^(24)`
The number of atoms of radioactive isotope left after one hour,
`N = (6.023 xx 10^(24) = 0.0003 xx 10^(24))`
`= 6.0227 xx 10^(24)`
Using, `lambda = (2.303)/(t) log (N_(0))/(N)`
`lambda = (2.303)/(t) log (6.023 xx 10^(24))/(6.0227 xx 10^(24))`
`= 2.303 xx 2.1632 xx 10^(-5) = 4.982 xx 10^(-5) hr^(-1)`
`t_(1//2) = .0693 // (4.982 xx 10^(-5) xx 24 xx 365) = 1.58` years