Suppose `3.0 xx 10^7` radon atoms are trapped in a basement at a given time. The basement is sealed against further enetry of the gas. The half - life of radon is 3.83 days. How many radon atoms remain after` 31` days?

Explanation

The number N of radon atoms remaining after a time t is given by `N =N_(0) e^(- lambda t)`, where `N_(0) = 3.0 xx 10^(7)` is the original number of atoms when `t =0 s` and `lambda` is the decay constant. The decay constant is related to the half-life `T_1//2` of the radon atoms by `lamda =0.639//T_(1)//2` . The activity can be obtained from the equation `Delta N//Delta t= - lambda N`.
(a) The decay constant is
`lambda =0.693/T_(1//2)=(0.693)/(3.83" "days) =0.181 days^(-1) =0.181 days^(-1)`
and the number N of radon atoms remaining after 31 days is
`N =N_(0) e^(- lambda t) =(3.0 xx 10^(7))exp{-(0.181 days^(-1)) (31 days)}`
`1.1 xx 10^(5)`
This value is slightly different from that found in Illustrartion `5.19` because there we ignored the difference between `8.0` and `8.1` half- lives.
(b) The acticity can be obtained from equation `Delta N//Delta t = - lambda N` , provided the decay constant is expressed in reciprocal second : `lambda =0.693//(3.31 xx 10^(5) s)=2.09 xx 10^(-6) s^(-1)`. Thus, the number of disintegrations per second is
`(Delta N)/(Delta t) = - lambda N= -(2.09 xx 10^(-6) s^(-1))(3.0 xx 10^(7)`
` = -63` disintergrations `s^(-1)`
The activity is the amgnitude of `Delta N//Delta t`, so initially Activity `= 63 Bq`
(c ) From part (a), the number of radioactive nuclei remaining at the end of `31` days is `N = 1.1 xx 10^(5)` , and reasoning similar to that in part (b) reveals that
Activity `=0.23 Bq`
SI unit of activity is becquerel (Bq) which is same as `1` dps (disintergrations per second). The popular unit of activity is cuires which is defined as
`1` curie `=3.7 xx 10^(10) dps` .