The activity of `1 g` radium is found to be 0.5. Calculate the half-life period of radium and the time required for the decay of `2 g` of radium to give `0.25 g` of radium (atomic mass off radium = 226).

Answer

9492yrs.

Explanation

Number of atoms in `1.0 g " of " .^(226)Ra=(6.023xx10^(23)xx1.0)/(226)`
`"Activity"-(dN)/(dt)=0.5 "curie"`
= `0.5xx3.7xx10^(10)` disintegrations per second
`=1.85xx10^(10)` disintegrations per second
`-(dN)/(dt)=lambdaN "or" 1.85xx10^(10)=lambdaxx(6.023xx10^(10))/(226)`
Or `lambda=6.945xx10^(-12)s^(-1)" Or " t_(1//2)=(0.693)/(6.945 xx10^(-12)) =9.978 xx10^(10)s`
`=(9.978xx10^(10))/(3600 xx24xx365) "years" =3164 "years"`
Time required for the decay of 2.0g to 0.25 g=three half lives=
`3 xx3164=9492`year