1.

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» Number of atoms in `1 g` of `Ra^(226) = (6.02 xx 10^(23))/(226)`
Activity, i.e.,
`(dN)/(dt) = 0.5` curie `= 0.5 xx 3.7 xx 10^(10) dps`
But `(dN)/(dt) = K//N`, i.e., `1.85 xx 10^(10) = K xx (6.02 xx 10^(23))/(226)`
or `K = 6.945 xx 10^(-12) s^(-1)`
`:. t_(1//2) = (0.693)/(K) = (0.693)/(6.945 xx 10^(-12)) s`
`= 9.978 xx 10^(10) s`
`= (9.978 xx 10^(10))/(3600 xx 24 xx 365)` years = 3164 years
Time requried for decay of `2 g` of `Ra` to `0.25 g` = three half lives `= 3 xx 3164 = 9492` years


Discussion

No Comment Found

Related InterviewSolutions