It is found that `3.125 xx 10^(-8) g` atoms of `Rn` exist in equilibrium with `1 g` of radium at `0^(@)C` and 1 atm pressure. The disintegration Constant of `Ra` is `1.48 xx 10^(11) s^(-1)`. Calculate the disintegration cosntant of `Rn`.

Explanation

Number of `g` atoms of `Ra^(226)` in `1 g` of `Ra^(226)`
`= 1//226 = 4.425 xx 10^(-3)`
At equilibrium `K_(A) N_(A) = K_(B) N_(B)`
`:. K_(Rn) xx N_(Rn) = K_(Ra) xx N_(Ra)`
or `(K_(Rn))/(K_(Ra)) = (N_(Ra))/(N_(Ra))`
`:. K_(Rn) = (N_(Ra))/(N_(Rn)) xx K_(Ra)`
`= (4.425 xx 10^(-3))/(3.125 xx 10^(-8)) xx 1.48 xx 10^(-11) s^(-1)`
`= 2.095 xx 10^(-6) s^(-1)`