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

`N_(0) = 10g` atoms `= 10xx6.023xx10^(230)` atoms
`= 6.023xx10^(24)` atoms
Volume of `He` collected `= 11.2 mL = (11.2)/(22400)` mol e
`= 5xx10^(-4) mol e = 5xx10^(-4) xx6.023xx10^(23)` atoms
`= 3.01xx10^(20)` atoms
The helium atoms formed
= No, of atoms of radioactive substace decacyed
`:.` No. of atoms of radioactive substance left `= N`
`= 6.023xx10^(24) - 3.01xx10^(20)`
`= 6.0227xx10^(24)` atoms
`:' lambda = (2.303)/(t) log_(10) (N_(0))/(N) = (2.303)/(1) log_(10) (6.023xx10^(24))/(6.0227xx10^(24))`
`= 4.982xx10^(-5) hr^(-1)`
`:. t_(1//2) = (0.693)/(lambda) = (0.693)/(4.982xx10^(-5))`
`= 13910.29` hour
Alternate solution:
Rate `= lambda-N`
Mole `"formed"//hr = "rate" = (11.2)/(22400)`
`:. (11.2)/(22400) = (0.693)/(t_(1//2)) xx 10`
`:. t_(1//2) = (0.693xx10xx22400)/(11.2) = 13860` hour.